Maximum possible difference of two subsets of an array
Last Updated :
24 Mar, 2023
Given an array of n-integers. The array may contain repetitive elements but the highest frequency of any element must not exceed two. You have to make two subsets such that the difference of the sum of their elements is maximum and both of them jointly contain all elements of the given array along with the most important condition, no subset should contain repetitive elements.
Examples:
Input : arr[] = {5, 8, -1, 4}
Output : Maximum Difference = 18
Explanation :
Let Subset A = {5, 8, 4} & Subset B = {-1}
Sum of elements of subset A = 17, of subset B = -1
Difference of Sum of Both subsets = 17 - (-1) = 18
Input : arr[] = {5, 8, 5, 4}
Output : Maximum Difference = 12
Explanation :
Let Subset A = {5, 8, 4} & Subset B = {5}
Sum of elements of subset A = 17, of subset B = 5
Difference of Sum of Both subsets = 17 - 5 = 12
Before solving this question we have to take care of some given conditions, and they are listed as:
- While building up the subsets, take care that no subset should contain repetitive elements. And for this, we can conclude that all such elements whose frequency are 2, going to be part of both subsets, and hence overall they don't have any impact on the difference of subset-sum. So, we can easily ignore them.
- For making the difference of the sum of elements of both subset maximum we have to make subset in such a way that all positive elements belong to one subset and negative ones to other subsets.
Algorithm with time complexity O(n2):
for i=0 to n-1
isSingleOccurrence = true;
for j= i+1 to n-1
// if frequency of any element is two
// make both equal to zero
if arr[i] equals arr[j]
arr[i] = arr[j] = 0
isSingleOccurrence = false;
break;
if isSingleOccurrence == true
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum2)
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++) {
bool isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++) {
// if frequency of any element is two
// make both equal to zero
if (arr[i] == arr[j]) {
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence) {
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// java find maximum difference
// of subset sum
import java .io.*;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++)
{
boolean isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
static public void main (String[] args)
{
int []arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.length;
System.out.println("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
# Python3 find maximum difference
# of subset sum
import math
# function for maximum subset diff
def maxDiff(arr, n) :
SubsetSum_1 = 0
SubsetSum_2 = 0
for i in range(0, n) :
isSingleOccurrence = True
for j in range(i + 1, n) :
# if frequency of any element
# is two make both equal to
# zero
if (arr[i] == arr[j]) :
isSingleOccurrence = False
arr[i] = arr[j] = 0
break
if (isSingleOccurrence == True) :
if (arr[i] > 0) :
SubsetSum_1 += arr[i]
else :
SubsetSum_2 += arr[i]
return abs(SubsetSum_1 - SubsetSum_2)
# Driver Code
arr = [4, 2, -3, 3, -2, -2, 8]
n = len(arr)
print ("Maximum Difference = {}"
. format(maxDiff(arr, n)))
# This code is contributed by Manish Shaw
# (manishshaw1)
// C# find maximum difference of
// subset sum
using System;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int SubsetSum_1 = 0, SubsetSum_2 = 0;
for (int i = 0; i <= n - 1; i++)
{
bool isSingleOccurrence = true;
for (int j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.Abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
static public void Main ()
{
int []arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.Length;
Console.WriteLine("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
<?php
// PHP find maximum difference
// of subset sum
// function for maximum subset diff
function maxDiff($arr, $n)
{
$SubsetSum_1 = 0;
$SubsetSum_2 = 0;
for ($i = 0; $i <= $n - 1; $i++)
{
$isSingleOccurrence = true;
for ($j = $i + 1; $j <= $n - 1; $j++)
{
// if frequency of any element is two
// make both equal to zero
if ($arr[$i] == $arr[$j])
{
$isSingleOccurrence = false;
$arr[$i] = $arr[$j] = 0;
break;
}
}
if ($isSingleOccurrence)
{
if ($arr[$i] > 0)
$SubsetSum_1 += $arr[$i];
else
$SubsetSum_2 += $arr[$i];
}
}
return abs($SubsetSum_1 - $SubsetSum_2);
}
// Driver Code
$arr = array(4, 2, -3, 3, -2, -2, 8);
$n = sizeof($arr);
echo "Maximum Difference = " , maxDiff($arr, $n);
// This code is contributed by nitin mittal
?>
<script>
// JavaScript Program to find maximum difference
// of subset sum
// function for maximum subset diff
function maxDiff(arr, n)
{
let SubsetSum_1 = 0, SubsetSum_2 = 0;
for (let i = 0; i <= n - 1; i++)
{
let isSingleOccurrence = true;
for (let j = i + 1; j <= n - 1; j++)
{
// if frequency of any element
// is two make both equal to
// zero
if (arr[i] == arr[j])
{
isSingleOccurrence = false;
arr[i] = arr[j] = 0;
break;
}
}
if (isSingleOccurrence)
{
if (arr[i] > 0)
SubsetSum_1 += arr[i];
else
SubsetSum_2 += arr[i];
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver program
let arr = [ 4, 2, -3, 3, -2, -2, 8 ];
let n = arr.length;
document.write("Maximum Difference = "
+ maxDiff(arr, n));
// This code is contributed by susmitakundugoaldanga.
</script>
OutputMaximum Difference = 20
Time Complexity O(n2)
Auxiliary Space: O(1)
Algorithm with time complexity O(n log n):
-> sort the array
-> for i =0 to n-2
// consecutive two elements are not equal
// add absolute arr[i] to result
if arr[i] != arr[i+1]
result += abs(arr[i])
// else skip next element too
else
i++;
// special check for last two elements
-> if (arr[n-2] != arr[n-1])
result += arr[n-1]
-> return result;
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
int result = 0;
// sort the array
sort(arr, arr + n);
// calculate the result
for (int i = 0; i < n - 1; i++) {
if (arr[i] != arr[i + 1])
result += abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += abs(arr[n - 1]);
// return result
return result;
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// java find maximum difference of
// subset sum
import java. io.*;
import java .util.*;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int result = 0;
// sort the array
Arrays.sort(arr);
// calculate the result
for (int i = 0; i < n - 1; i++)
{
if (arr[i] != arr[i + 1])
result += Math.abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.abs(arr[n - 1]);
// return result
return result;
}
// driver program
static public void main (String[] args)
{
int[] arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.length;
System.out.println("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
# Python 3 find maximum difference
# of subset sum
# function for maximum subset diff
def maxDiff(arr, n):
result = 0
# sort the array
arr.sort()
# calculate the result
for i in range(n - 1):
if (abs(arr[i]) != abs(arr[i + 1])):
result += abs(arr[i])
else:
pass
# check for last element
if (arr[n - 2] != arr[n - 1]):
result += abs(arr[n - 1])
# return result
return result
# Driver Code
if __name__ == "__main__":
arr = [ 4, 2, -3, 3, -2, -2, 8 ]
n = len(arr)
print("Maximum Difference = " ,
maxDiff(arr, n))
# This code is contributed by ita_c
// C# find maximum difference
// of subset sum
using System;
public class GFG {
// function for maximum subset diff
static int maxDiff(int []arr, int n)
{
int result = 0;
// sort the array
Array.Sort(arr);
// calculate the result
for (int i = 0; i < n - 1; i++)
{
if (arr[i] != arr[i + 1])
result += Math.Abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.Abs(arr[n - 1]);
// return result
return result;
}
// driver program
static public void Main ()
{
int[] arr = { 4, 2, -3, 3, -2, -2, 8 };
int n = arr.Length;
Console.WriteLine("Maximum Difference = "
+ maxDiff(arr, n));
}
}
// This code is contributed by vt_m.
<?php
// PHP find maximum difference of subset sum
// function for maximum subset diff
function maxDiff( $arr, $n)
{
$result = 0;
// sort the array
sort($arr);
// calculate the result
for ( $i = 0; $i < $n - 1; $i++)
{
if ($arr[$i] != $arr[$i + 1])
$result += abs($arr[$i]);
else
$i++;
}
// check for last element
if ($arr[$n - 2] != $arr[$n - 1])
$result += abs($arr[$n - 1]);
// return result
return $result;
}
// Driver Code
$arr = array( 4, 2, -3, 3, -2, -2, 8 );
$n = count($arr);
echo "Maximum Difference = "
, maxDiff($arr, $n);
// This code is contributed by anuj_67.
?>
<script>
// Javascript find maximum difference of subset sum
// function for maximum subset diff
function maxDiff(arr, n)
{
var result = 0;
// sort the array
arr.sort((a,b)=> a-b)
// calculate the result
for (var i = 0; i < n - 1; i++) {
if (arr[i] != arr[i + 1])
result += Math.abs(arr[i]);
else
i++;
}
// check for last element
if (arr[n - 2] != arr[n - 1])
result += Math.abs(arr[n - 1]);
// return result
return result;
}
// driver program
var arr = [ 4, 2, -3, 3, -2, -2, 8 ];
var n = arr.length;
document.write( "Maximum Difference = " + maxDiff(arr, n));
</script>
OutputMaximum Difference = 20
Time Complexity: O(n log n)
Auxiliary Space: O(1)
Algorithm with time complexity O(n):
make hash table for positive elements:
for all positive elements(arr[i])
if frequency == 1
SubsetSum_1 += arr[i];
make hash table for negative elements:
for all negative elements
if frequency == 1
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum2)
Implementation:
C++
// CPP find maximum difference of subset sum
#include <bits/stdc++.h>
using namespace std;
// function for maximum subset diff
int maxDiff(int arr[], int n)
{
unordered_map<int, int> hashPositive;
unordered_map<int, int> hashNegative;
int SubsetSum_1 = 0, SubsetSum_2 = 0;
// construct hash for positive elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] > 0)
hashPositive[arr[i]]++;
// calculate subset sum for positive elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] > 0 && hashPositive[arr[i]] == 1)
SubsetSum_1 += arr[i];
// construct hash for negative elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] < 0)
hashNegative[abs(arr[i])]++;
// calculate subset sum for negative elements
for (int i = 0; i <= n - 1; i++)
if (arr[i] < 0 &&
hashNegative[abs(arr[i])] == 1)
SubsetSum_2 += arr[i];
return abs(SubsetSum_1 - SubsetSum_2);
}
// driver program
int main()
{
int arr[] = { 4, 2, -3, 3, -2, -2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum Difference = " << maxDiff(arr, n);
return 0;
}
// Java find maximum
// difference of subset sum
import java.util.*;
class GFG{
// Function for maximum subset diff
public static int maxDiff(int arr[],
int n)
{
HashMap<Integer,
Integer> hashPositive = new HashMap<>();
HashMap<Integer,
Integer> hashNegative = new HashMap<>();
int SubsetSum_1 = 0,
SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.containsKey(arr[i]))
{
hashPositive.replace(arr[i],
hashPositive.get(arr[i]) + 1);
}
else
{
hashPositive.put(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (int i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 &&
hashPositive.containsKey(arr[i]))
{
if(hashPositive.get(arr[i]) == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.containsKey(Math.abs(arr[i])))
{
hashNegative.replace(Math.abs(arr[i]),
hashNegative.get(Math.abs(arr[i])) + 1);
}
else
{
hashNegative.put(Math.abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.containsKey(Math.abs(arr[i])))
{
if(hashNegative.get(Math.abs(arr[i])) == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
public static void main(String[] args)
{
int arr[] = {4, 2, -3, 3,
-2, -2, 8};
int n = arr.length;
System.out.print("Maximum Difference = " +
maxDiff(arr, n));
}
}
// This code is contributed by divyeshrabadiya07
# Python3 find maximum difference of subset sum
# function for maximum subset diff
def maxDiff(arr, n):
hashPositive = dict()
hashNegative = dict()
SubsetSum_1, SubsetSum_2 = 0, 0
# construct hash for positive elements
for i in range(n):
if (arr[i] > 0):
hashPositive[arr[i]] = \
hashPositive.get(arr[i], 0) + 1
# calculate subset sum for positive elements
for i in range(n):
if (arr[i] > 0 and arr[i] in
hashPositive.keys() and
hashPositive[arr[i]] == 1):
SubsetSum_1 += arr[i]
# construct hash for negative elements
for i in range(n):
if (arr[i] < 0):
hashNegative[abs(arr[i])] = \
hashNegative.get(abs(arr[i]), 0) + 1
# calculate subset sum for negative elements
for i in range(n):
if (arr[i] < 0 and abs(arr[i]) in
hashNegative.keys() and
hashNegative[abs(arr[i])] == 1):
SubsetSum_2 += arr[i]
return abs(SubsetSum_1 - SubsetSum_2)
# Driver Code
arr = [4, 2, -3, 3, -2, -2, 8]
n = len(arr)
print("Maximum Difference =", maxDiff(arr, n))
# This code is contributed by mohit kumar
// C# find maximum
// difference of subset sum
using System;
using System.Collections.Generic;
class GFG {
// Function for maximum subset diff
static int maxDiff(int[] arr, int n)
{
Dictionary<int, int> hashPositive =
new Dictionary<int, int>();
Dictionary<int, int> hashNegative =
new Dictionary<int, int>();
int SubsetSum_1 = 0, SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.ContainsKey(arr[i]))
{
hashPositive[arr[i]] += 1;
}
else
{
hashPositive.Add(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (int i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 && hashPositive.ContainsKey(arr[i]))
{
if(hashPositive[arr[i]] == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.ContainsKey(Math.Abs(arr[i])))
{
hashNegative[(Math.Abs(arr[i]))] += 1;
}
else
{
hashNegative.Add(Math.Abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (int i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.ContainsKey(Math.Abs(arr[i])))
{
if(hashNegative[(Math.Abs(arr[i]))] == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.Abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
static void Main() {
int[] arr = {4, 2, -3, 3, -2, -2, 8};
int n = arr.Length;
Console.WriteLine("Maximum Difference = " +
maxDiff(arr, n));
}
}
// This code is contributed by divesh072019
<script>
// Javascript find maximum
// difference of subset sum
// Function for maximum subset diff
function maxDiff(arr,n)
{
let hashPositive = new Map();
let hashNegative = new Map();
let SubsetSum_1 = 0,
SubsetSum_2 = 0;
// Construct hash for
// positive elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] > 0)
{
if(hashPositive.has(arr[i]))
{
hashPositive.set(arr[i],
hashPositive.get(arr[i]) + 1);
}
else
{
hashPositive.set(arr[i], 1);
}
}
}
// Calculate subset sum
// for positive elements
for (let i = 0; i <= n - 1; i++)
{
if(arr[i] > 0 &&
hashPositive.has(arr[i]))
{
if(hashPositive.get(arr[i]) == 1)
{
SubsetSum_1 += arr[i];
}
}
}
// Construct hash for
// negative elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] < 0)
{
if(hashNegative.has(Math.abs(arr[i])))
{
hashNegative.set(Math.abs(arr[i]),
hashNegative.get(Math.abs(arr[i])) + 1);
}
else
{
hashNegative.set(Math.abs(arr[i]), 1);
}
}
}
// Calculate subset sum for
// negative elements
for (let i = 0; i <= n - 1; i++)
{
if (arr[i] < 0 &&
hashNegative.has(Math.abs(arr[i])))
{
if(hashNegative.get(Math.abs(arr[i])) == 1)
{
SubsetSum_2 += arr[i];
}
}
}
return Math.abs(SubsetSum_1 - SubsetSum_2);
}
// Driver code
let arr = [4, 2, -3, 3,
-2, -2, 8];
let n = arr.length;
document.write("Maximum Difference = " +
maxDiff(arr, n));
// This code is contributed by rag2127
</script>
OutputMaximum Difference = 20
Time Complexity: O(n)
Auxiliary Space: O(n)
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Union and Intersection of two Linked List using HashingGiven two singly Linked Lists, create union and intersection lists that contain the union and intersection of the elements present in the given lists. Each of the two linked lists contains distinct node values.Note: The order of elements in output lists doesn’t matter. Examples:Input:head1 : 10 -
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Two Sum - Pair with given SumGiven an array arr[] of n integers and a target value, the task is to find whether there is a pair of elements in the array whose sum is equal to target. This problem is a variation of 2Sum problem.Examples: Input: arr[] = [0, -1, 2, -3, 1], target = -2Output: trueExplanation: There is a pair (1, -3
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Max Distance Between Two OccurrencesGiven an array arr[], the task is to find the maximum distance between two occurrences of any element. If no element occurs twice, return 0.Examples: Input: arr = [1, 1, 2, 2, 2, 1]Output: 5Explanation: distance for 1 is: 5-0 = 5, distance for 2 is: 4-2 = 2, So max distance is 5.Input : arr[] = [3,
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Most frequent element in an arrayGiven an array, the task is to find the most frequent element in it. If there are multiple elements that appear a maximum number of times, return the maximum element.Examples: Input : arr[] = [1, 3, 2, 1, 4, 1]Output : 1Explanation: 1 appears three times in array which is maximum frequency.Input : a
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Only Repeating From 1 To n-1Given an array arr[] of size n filled with numbers from 1 to n-1 in random order. The array has only one repetitive element. The task is to find the repetitive element.Examples:Input: arr[] = [1, 3, 2, 3, 4]Output: 3Explanation: The number 3 is the only repeating element.Input: arr[] = [1, 5, 1, 2,
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Check for Disjoint Arrays or SetsGiven two arrays a and b, check if they are disjoint, i.e., there is no element common between both the arrays.Examples: Input: a[] = {12, 34, 11, 9, 3}, b[] = {2, 1, 3, 5} Output: FalseExplanation: 3 is common in both the arrays.Input: a[] = {12, 34, 11, 9, 3}, b[] = {7, 2, 1, 5} Output: True Expla
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Non-overlapping sum of two setsGiven two arrays A[] and B[] of size n. It is given that both array individually contains distinct elements. We need to find the sum of all elements that are not common.Examples: Input : A[] = {1, 5, 3, 8} B[] = {5, 4, 6, 7}Output : 291 + 3 + 4 + 6 + 7 + 8 = 29Input : A[] = {1, 5, 3, 8} B[] = {5, 1,
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Check if two arrays are equal or notGiven two arrays, a and b of equal length. The task is to determine if the given arrays are equal or not. Two arrays are considered equal if:Both arrays contain the same set of elements.The arrangements (or permutations) of elements may be different.If there are repeated elements, the counts of each
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Find missing elements of a rangeGiven an array, arr[0..n-1] of distinct elements and a range [low, high], find all numbers that are in a range, but not the array. The missing elements should be printed in sorted order.Examples: Input: arr[] = {10, 12, 11, 15}, low = 10, high = 15Output: 13, 14Input: arr[] = {1, 14, 11, 51, 15}, lo
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Minimum Subsets with Distinct ElementsYou are given an array of n-element. You have to make subsets from the array such that no subset contain duplicate elements. Find out minimum number of subset possible.Examples : Input : arr[] = {1, 2, 3, 4}Output :1Explanation : A single subset can contains all values and all values are distinct.In
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Remove minimum elements such that no common elements exist in two arraysGiven two arrays arr1[] and arr2[] consisting of n and m elements respectively. The task is to find the minimum number of elements to remove from each array such that intersection of both arrays becomes empty and both arrays become mutually exclusive.Examples: Input: arr[] = { 1, 2, 3, 4}, arr2[] =
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2 Sum - Count pairs with given sumGiven an array arr[] of n integers and a target value, the task is to find the number of pairs of integers in the array whose sum is equal to target.Examples: Input: arr[] = {1, 5, 7, -1, 5}, target = 6Output: 3Explanation: Pairs with sum 6 are (1, 5), (7, -1) & (1, 5). Input: arr[] = {1, 1, 1,
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Count quadruples from four sorted arrays whose sum is equal to a given value xGiven four sorted arrays each of size n of distinct elements. Given a value x. The problem is to count all quadruples(group of four numbers) from all the four arrays whose sum is equal to x.Note: The quadruple has an element from each of the four arrays. Examples: Input : arr1 = {1, 4, 5, 6}, arr2 =
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Sort elements by frequency | Set 4 (Efficient approach using hash)Print the elements of an array in the decreasing frequency if 2 numbers have the same frequency then print the one which came first. Examples: Input : arr[] = {2, 5, 2, 8, 5, 6, 8, 8} Output : arr[] = {8, 8, 8, 2, 2, 5, 5, 6} Input : arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8} Output : arr[] = {8,
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Find all pairs (a, b) in an array such that a % b = kGiven an array with distinct elements, the task is to find the pairs in the array such that a % b = k, where k is a given integer. You may assume that a and b are in small range Examples : Input : arr[] = {2, 3, 5, 4, 7} k = 3Output : (7, 4), (3, 4), (3, 5), (3, 7)7 % 4 = 33 % 4 = 33 % 5 = 33 % 7 =
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Group words with same set of charactersGiven a list of words with lower cases. Implement a function to find all Words that have the same unique character set. Example: Input: words[] = { "may", "student", "students", "dog", "studentssess", "god", "cat", "act", "tab", "bat", "flow", "wolf", "lambs", "amy", "yam", "balms", "looped", "poodl
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k-th distinct (or non-repeating) element among unique elements in an array.Given an integer array arr[], print kth distinct element in this array. The given array may contain duplicates and the output should print the k-th element among all unique elements. If k is more than the number of distinct elements, print -1.Examples:Input: arr[] = {1, 2, 1, 3, 4, 2}, k = 2Output:
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Intermediate problems on Hashing
Find Itinerary from a given list of ticketsGiven a list of tickets, find the itinerary in order using the given list.Note: It may be assumed that the input list of tickets is not cyclic and there is one ticket from every city except the final destination.Examples:Input: "Chennai" -> "Bangalore" "Bombay" -> "Delhi" "Goa" -> "Chennai"
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Find number of Employees Under every ManagerGiven a 2d matrix of strings arr[][] of order n * 2, where each array arr[i] contains two strings, where the first string arr[i][0] is the employee and arr[i][1] is his manager. The task is to find the count of the number of employees under each manager in the hierarchy and not just their direct rep
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Longest Subarray With Sum Divisible By KGiven an arr[] containing n integers and a positive integer k, he problem is to find the longest subarray's length with the sum of the elements divisible by k.Examples:Input: arr[] = [2, 7, 6, 1, 4, 5], k = 3Output: 4Explanation: The subarray [7, 6, 1, 4] has sum = 18, which is divisible by 3.Input:
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Longest Subarray with 0 Sum Given an array arr[] of size n, the task is to find the length of the longest subarray with sum equal to 0.Examples:Input: arr[] = {15, -2, 2, -8, 1, 7, 10, 23}Output: 5Explanation: The longest subarray with sum equals to 0 is {-2, 2, -8, 1, 7}Input: arr[] = {1, 2, 3}Output: 0Explanation: There is n
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Longest Increasing consecutive subsequenceGiven N elements, write a program that prints the length of the longest increasing consecutive subsequence. Examples: Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12} Output : 6 Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one. Input : a[] = {6
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Count Distinct Elements In Every Window of Size KGiven an array arr[] of size n and an integer k, return the count of distinct numbers in all windows of size k. Examples: Input: arr[] = [1, 2, 1, 3, 4, 2, 3], k = 4Output: [3, 4, 4, 3]Explanation: First window is [1, 2, 1, 3], count of distinct numbers is 3. Second window is [2, 1, 3, 4] count of d
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Design a data structure that supports insert, delete, search and getRandom in constant timeDesign a data structure that supports the following operations in O(1) time.insert(x): Inserts an item x to the data structure if not already present.remove(x): Removes item x from the data structure if present. search(x): Searches an item x in the data structure.getRandom(): Returns a random elemen
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Subarray with Given Sum - Handles Negative NumbersGiven an unsorted array of integers, find a subarray that adds to a given number. If there is more than one subarray with the sum of the given number, print any of them.Examples: Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33Output: Sum found between indexes 2 and 4Explanation: Sum of elements betwee
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Implementing our Own Hash Table with Separate Chaining in JavaAll data structure has their own special characteristics, for example, a BST is used when quick searching of an element (in log(n)) is required. A heap or a priority queue is used when the minimum or maximum element needs to be fetched in constant time. Similarly, a hash table is used to fetch, add
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Implementing own Hash Table with Open Addressing Linear ProbingIn Open Addressing, all elements are stored in the hash table itself. So at any point, size of table must be greater than or equal to total number of keys (Note that we can increase table size by copying old data if needed).Insert(k) - Keep probing until an empty slot is found. Once an empty slot is
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Maximum possible difference of two subsets of an arrayGiven an array of n-integers. The array may contain repetitive elements but the highest frequency of any element must not exceed two. You have to make two subsets such that the difference of the sum of their elements is maximum and both of them jointly contain all elements of the given array along w
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Sorting using trivial hash functionWe have read about various sorting algorithms such as heap sort, bubble sort, merge sort and others. Here we will see how can we sort N elements using a hash array. But this algorithm has a limitation. We can sort only those N elements, where the value of elements is not large (typically not above 1
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Smallest subarray with k distinct numbersWe are given an array consisting of n integers and an integer k. We need to find the smallest subarray [l, r] (both l and r are inclusive) such that there are exactly k different numbers. If no such subarray exists, print -1 and If multiple subarrays meet the criteria, return the one with the smalle
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