Cost of creating smallest subsequence with sum of difference between adjacent elements maximum
Last Updated :
05 Sep, 2022
Given two array of N integers arr[] and costArray[], representing removal cost associated with each element. The task is to find the subsequence from the given array of minimum cost such that the sum of the difference between adjacent elements is maximum. On removing each element, cost is incurred.
Examples:
Input: N = 3, arr[] = { 3, 2, 1 }, costArray[] = {0, 1, 0}
Output: 4, 1
Explanation:
There are 4 subsequences of length at least 2 :
[ 3, 2 ] which gives us | 3 - 2 | = 1
[ 3, 1 ] which gives us | 3 - 1 | = 2
[ 2, 1 ] which gives us | 2 - 1 | = 1
[ 3, 2, 1 ] which gives us | 3 - 2 | + | 2 - 1 | = 2
So the answer is either [ 3, 1 ] or [ 3, 2, 1 ] . Since we want the subsequence to be as short as possible, the answer is [ 3, 1 ]. Cost incurrent removing element 2 is 1.
So, the sum of the sequence is 4.
and the cost is 1.
Input: N = 4, arr[] = { 1, 3, 4, 2}, costArray[] = {0, 1, 0, 0}
Output: 7, 1
Naive Approach:
The naive approach is simply to check all the subsequences by generating them by recursion and it takes the complexity of O(2^N) which is very high complexity. And from them choose the subsequence which follows the above condition with maximum absolute sum and a minimum length as mentioned above.
Efficient Approach:
We can observe the pattern, let us assume three numbers a, b, c such that a = L, b = L + 6, c = L + 10 (L is any integer here). If they are in a continuous pattern one after another eg a, b, c (such that a < b < c).- Then
| b – a | + | c – b | = | a – c | = 10
here we can remove the middle element to reduce the size of the original sequence.- In this way, reducing the size of the array by removing a middle element from the sequence will not affect the sum and also reduce the length of the sequence.
- Stores all the removed elements in the set. Add the cost of removed elements. Finally, calculate the sum sequence by excluding removed elements.
- Then print the sum of elements of subsequence and cost incurred.
In this way, we reduce the exponential complexity O(2^N) to the linear complexity O(N).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void costOfSubsequence(
int N, int arr[],
int costArray[])
{
int i, temp;
// initializing cost=0
int cost = 0;
// to store the removed
// element
set<int> removedElements;
// this will store the sum
// of the subsequence
int ans = 0;
// checking all the element
// of the vector
for (i = 1; i < (N - 1); i++) {
// storing the value of
// arr[i] in temp variable
temp = arr[i];
// if the situation like
// arr[i-1]<arr[i]<arr[i+1] or
// arr[i-1]>arr[i]>arr[i+1] occur
// remove arr[i] i.e, temp
// from sequence
if (((arr[i - 1] < temp)
&& (temp < arr[i + 1]))
|| ((arr[i - 1] > temp)
&& (temp > arr[i + 1]))) {
// insert the element in the set
// removedElements
removedElements.insert(temp);
}
}
for (i = 0; i < (N); i++) {
// storing the value of
// arr[i] in temp
temp = arr[i];
// taking the element not
// in removedElements
if (!(removedElements.count(temp) > 0)) {
// adding the value of elements
// of subsequence
ans += arr[i];
}
else {
// if we have to remove
// the element then we
// need to add the cost
// associated with the
// element
cost += costArray[i];
}
}
// printing the sum of
// the subsequence with
// minimum length possible
cout << ans << ", ";
// printing the cost incurred
// in creating subsequence
cout << cost << endl;
}
// Driver code
int main()
{
int N;
N = 4;
int arr[N]
= { 1, 3, 4, 2 };
int costArray[N]
= { 0, 1, 0, 0 };
// calling the function
costOfSubsequence(
N, arr,
costArray);
return 0;
}
import java.util.*;
class GFG{
public static void costOfSubsequence(int N, int[] arr,
int[] costArray)
{
int i, temp;
// Initializing cost=0
int cost = 0;
// To store the removed
// element
Set<Integer> removedElements = new HashSet<Integer>();
// This will store the sum
// of the subsequence
int ans = 0;
// Checking all the element
// of the vector
for(i = 1; i < (N - 1); i++)
{
// Storing the value of
// arr[i] in temp variable
temp = arr[i];
// If the situation like
// arr[i-1]<arr[i]<arr[i+1] or
// arr[i-1]>arr[i]>arr[i+1] occur
// remove arr[i] i.e, temp
// from sequence
if (((arr[i - 1] < temp) &&
(temp < arr[i + 1])) ||
((arr[i - 1] > temp) &&
(temp > arr[i + 1])))
{
// Insert the element in the set
// removedElements
removedElements.add(temp);
}
}
for(i = 0; i < (N); i++)
{
// Storing the value of
// arr[i] in temp
temp = arr[i];
// Taking the element not
// in removedElements
if (!(removedElements.contains(temp)))
{
// Adding the value of elements
// of subsequence
ans += arr[i];
}
else
{
// If we have to remove
// the element then we
// need to add the cost
// associated with the
// element
cost += costArray[i];
}
}
// Printing the sum of
// the subsequence with
// minimum length possible
System.out.print(ans + ", ");
// Printing the cost incurred
// in creating subsequence
System.out.print(cost);
}
// Driver code
public static void main(String[] args)
{
int N;
N = 4;
int[] arr = { 1, 3, 4, 2 };
int[] costArray = { 0, 1, 0, 0 };
// Calling the function
costOfSubsequence(N, arr, costArray);
}
}
// This code is contributed by divyeshrabadiya07
def costOfSubsequence(N, arr, costArray):
# Initializing cost=0
i, temp, cost = 0, 0, 0
# To store the removed
# element
removedElements = {''}
# This will store the sum
# of the subsequence
ans = 0
# Checking all the element
# of the vector
for i in range(1, N - 1):
# Storing the value of
# arr[i] in temp variable
temp = arr[i]
# If the situation like
# arr[i-1]<arr[i]<arr[i+1] or
# arr[i-1]>arr[i]>arr[i+1] occur
# remove arr[i] i.e, temp
# from sequence
if (((arr[i - 1] < temp) and
(temp < arr[i + 1])) or
((arr[i - 1] > temp) and
(temp > arr[i + 1]))) :
# Insert the element in the set
# removedElements
removedElements.add(temp)
for i in range(0, N):
# Storing the value of
# arr[i] in temp
temp = arr[i]
# Taking the element not
# in removedElements
if(temp not in removedElements):
# Adding the value of elements
# of subsequence
ans = ans + arr[i]
else:
# If we have to remove
# the element then we
# need to add the cost
# associated with the
# element
cost += costArray[i]
# Printing the sum of
# the subsequence with
# minimum length possible
print(ans, end = ", ")
# Printing the cost incurred
# in creating subsequence
print(cost)
# Driver code
N = 4
arr = [ 1, 3, 4, 2 ]
costArray = [ 0, 1, 0, 0 ]
# Calling the function
costOfSubsequence(N, arr, costArray)
# This code is contributed by Sanjit_Prasad
using System;
using System.Collections.Generic;
class GFG{
public static void costOfSubsequence(int N, int[] arr,
int[] costArray)
{
int i, temp;
// Initializing cost=0
int cost = 0;
// To store the removed
// element
HashSet<int> removedElements = new HashSet<int>();
// This will store the sum
// of the subsequence
int ans = 0;
// Checking all the element
// of the vector
for(i = 1; i < (N - 1); i++)
{
// Storing the value of
// arr[i] in temp variable
temp = arr[i];
// If the situation like
// arr[i-1]<arr[i]<arr[i+1] or
// arr[i-1]>arr[i]>arr[i+1] occur
// remove arr[i] i.e, temp
// from sequence
if (((arr[i - 1] < temp) &&
(temp < arr[i + 1])) ||
((arr[i - 1] > temp) &&
(temp > arr[i + 1])))
{
// Insert the element in the set
// removedElements
removedElements.Add(temp);
}
}
for(i = 0; i < (N); i++)
{
// Storing the value of
// arr[i] in temp
temp = arr[i];
// Taking the element not
// in removedElements
if (!(removedElements.Contains(temp)))
{
// Adding the value of elements
// of subsequence
ans += arr[i];
}
else
{
// If we have to remove
// the element then we
// need to add the cost
// associated with the
// element
cost += costArray[i];
}
}
// Printing the sum of
// the subsequence with
// minimum length possible
Console.Write(ans + ", ");
// Printing the cost incurred
// in creating subsequence
Console.Write(cost);
}
// Driver code
public static void Main(String[] args)
{
int N;
N = 4;
int[] arr = { 1, 3, 4, 2 };
int[] costArray = { 0, 1, 0, 0 };
// Calling the function
costOfSubsequence(N, arr, costArray);
}
}
// This code is contributed by Amit Katiyar
<script>
function costOfSubsequence(N,arr,costArray)
{
let i, temp;
// initializing cost=0
let cost = 0;
// to store the removed
// element
let removedElements = new Set();
// this will store the sum
// of the subsequence
let ans = 0;
// checking all the element
// of the vector
for (i = 1; i < (N - 1); i++) {
// storing the value of
// arr[i] in temp variable
temp = arr[i];
// if the situation like
// arr[i-1]<arr[i]<arr[i+1] or
// arr[i-1]>arr[i]>arr[i+1] occur
// remove arr[i] i.e, temp
// from sequence
if (((arr[i - 1] < temp)
&& (temp < arr[i + 1]))
|| ((arr[i - 1] > temp)
&& (temp > arr[i + 1]))) {
// insert the element in the set
// removedElements
removedElements.add(temp);
}
}
for (i = 0; i < (N); i++) {
// storing the value of
// arr[i] in temp
temp = arr[i];
// taking the element not
// in removedElements
if (!removedElements.has(temp)) {
// adding the value of elements
// of subsequence
ans += arr[i];
}
else {
// if we have to remove
// the element then we
// need to add the cost
// associated with the
// element
cost += costArray[i];
}
}
// printing the sum of
// the subsequence with
// minimum length possible
document.write(ans + ", ");
// printing the cost incurred
// in creating subsequence
document.write(cost);
}
// Driver code
let N;
N = 4;
let arr= [ 1, 3, 4, 2 ];
let costArray= [ 0, 1, 0, 0 ];
// calling the function
costOfSubsequence(N, arr,costArray);
</script>
Time Complexity: O(NlogN)
Auxiliary Space: O(N) because it uses extra space for set removedElements
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