Smallest subarray with all occurrences of a most frequent element
Last Updated :
22 Dec, 2022
Given an array, A. Let x be an element in the array. x has the maximum frequency in the array. Find the smallest subsegment of the array which also has x as the maximum frequency element.
Examples:
Input : arr[] = {4, 1, 1, 2, 2, 1, 3, 3}
Output : 1, 1, 2, 2, 1
The most frequent element is 1. The smallest
subarray that has all occurrences of it is
1 1 2 2 1
Input : A[] = {1, 2, 2, 3, 1}
Output : 2, 2
Note that there are two elements that appear
two times, 1 and 2. The smallest window for
1 is whole array and smallest window for 2 is
{2, 2}. Since window for 2 is smaller, this is
our output.
Approach:
Observe that if X is the maximum repeated element of our subsegment then the subsegment should look like this [X, ....., X], cause if the subsegment end or begins with another element we can delete it which does not alter our answer.
To solve this problem, let us store for every distinct element in the array three values, index of the first occurrence of the element and the index of the last occurrence the element and the frequency of the element. And at every step for a maximum repeated element minimize the size of our subsegment.
C++
// C++ implementation to find smallest
// subarray with all occurrences of
// a most frequent element
#include <bits/stdc++.h>
using namespace std;
void smallestSubsegment(int a[], int n)
{
// To store left most occurrence of elements
unordered_map<int, int> left;
// To store counts of elements
unordered_map<int, int> count;
// To store maximum frequency
int mx = 0;
// To store length and starting index of
// smallest result window
int mn, strindex;
for (int i = 0; i < n; i++) {
int x = a[i];
// First occurrence of an element,
// store the index
if (count[x] == 0) {
left[x] = i;
count[x] = 1;
}
// increase the frequency of elements
else
count[x]++;
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count[x] > mx) {
mx = count[x];
mn = i - left[x] + 1; // length of subsegment
strindex = left[x];
}
// select subsegment of smallest size
else if (count[x] == mx && i - left[x] + 1 < mn) {
mn = i - left[x] + 1;
strindex = left[x];
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = strindex; i < strindex + mn; i++)
cout << a[i] << " ";
}
// Driver code
int main()
{
int A[] = { 1, 2, 2, 2, 1 };
int n = sizeof(A) / sizeof(A[0]);
smallestSubsegment(A, n);
return 0;
}
// Java implementation to find smallest
// subarray with all occurrences of
// a most frequent element
import java.io.*;
import java.util.*;
class GfG {
static class Pair{
int fre;
int fi;
int li;
Pair(int fre,int fi,int li){
this.fre=fre;
this.fi=fi;
this.li=li;
}
}
static void smallestSubsegment(int arr[], int n)
{
// To store left most occurrence of elements
HashMap<Integer,Pair>hm=new HashMap<>();
int mfi=arr[0];
int mf=1;
int si=0;
int mflen=1;
hm.put(arr[0],new Pair(1,0,0));
for(int i=1;i<arr.length;i++){
int val=arr[i];
if(hm.containsKey(val))
{
Pair p=hm.get(val);
p.fre++;
p.li=i;
int len=i-p.fi+1;
if(p.fre>mf){
mfi=val;
mf=p.fre;
si=p.fi;
mflen=len;
}
else if(p.fre==mf && len<mflen){
mfi=val;
mf=p.fre;
si=p.fi;
mflen=len;
}
else if(p.fre==mf && len==mflen && p.fi<si)
{
mfi=val;
mf=p.fre;
si=p.fi;
mflen=len;
}
}
else
hm.put(val,new Pair(1,i,i));
}
int en=mflen+si-1;
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = si; i <=en ; i++)
System.out.print(arr[i] + " ");
}
// Driver program
public static void main (String[] args)
{
int A[] = { 1, 2, 2, 2, 1 };
int n = A.length;
smallestSubsegment(A, n);
}
}
// This code is contributed by waris amir
# Python3 implementation to find smallest
# subarray with all occurrences of
# a most frequent element
def smallestSubsegment(a, n):
# To store left most occurrence of elements
left = dict()
# To store counts of elements
count = dict()
# To store maximum frequency
mx = 0
# To store length and starting index of
# smallest result window
mn, strindex = 0, 0
for i in range(n):
x = a[i]
# First occurrence of an element,
# store the index
if (x not in count.keys()):
left[x] = i
count[x] = 1
# increase the frequency of elements
else:
count[x] += 1
# Find maximum repeated element and
# store its last occurrence and first
# occurrence
if (count[x] > mx):
mx = count[x]
mn = i - left[x] + 1 # length of subsegment
strindex = left[x]
# select subsegment of smallest size
elif (count[x] == mx and
i - left[x] + 1 < mn):
mn = i - left[x] + 1
strindex = left[x]
# Print the subsegment with all occurrences of
# a most frequent element
for i in range(strindex, strindex + mn):
print(a[i], end = " ")
# Driver code
A = [1, 2, 2, 2, 1]
n = len(A)
smallestSubsegment(A, n)
# This code is contributed by Mohit Kumar
// C# implementation to find smallest
// subarray with all occurrences of
// a most frequent element
using System;
using System.Collections.Generic;
class GfG
{
static void smallestSubsegment(int []a, int n)
{
// To store left most occurrence of elements
Dictionary<int, int> left = new Dictionary<int, int>();
// To store counts of elements
Dictionary<int, int> count = new Dictionary<int, int>();
// To store maximum frequency
int mx = 0;
// To store length and starting index of
// smallest result window
int mn = -1, strindex = -1;
for (int i = 0; i < n; i++)
{
int x = a[i];
// First occurrence of an element,
// store the index
if (!count.ContainsKey(x))
{
left.Add(x, i) ;
count.Add(x, 1);
}
// increase the frequency of elements
else
count[x] = count[x] + 1;
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count[x] > mx)
{
mx = count[x];
// length of subsegment
mn = i - left[x] + 1;
strindex = left[x];
}
// select subsegment of smallest size
else if ((count[x] == mx) &&
(i - left[x] + 1 < mn))
{
mn = i - left[x] + 1;
strindex = left[x];
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = strindex; i < strindex + mn; i++)
Console.Write(a[i] + " ");
}
// Driver code
public static void Main (String[] args)
{
int []A = { 1, 2, 2, 2, 1 };
int n = A.Length;
smallestSubsegment(A, n);
}
}
// This code is contributed by PrinciRaj1992
<script>
// JavaScript implementation to find smallest
// subarray with all occurrences of
// a most frequent element
function smallestSubsegment(a,n)
{
// To store left most occurrence of elements
let left= new Map();
// To store counts of elements
let count= new Map();
// To store maximum frequency
let mx = 0;
// To store length and starting index of
// smallest result window
let mn = -1, strindex = -1;
for (let i = 0; i < n; i++)
{
let x = a[i];
// First occurrence of an element,
// store the index
if (count.get(x) == null)
{
left.set(x, i) ;
count.set(x, 1);
}
// increase the frequency of elements
else
count.set(x, count.get(x) + 1);
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count.get(x) > mx)
{
mx = count.get(x);
// length of subsegment
mn = i - left.get(x) + 1;
strindex = left.get(x);
}
// select subsegment of smallest size
else if ((count.get(x) == mx) &&
(i - left.get(x) + 1 < mn))
{
mn = i - left.get(x) + 1;
strindex = left.get(x);
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (let i = strindex; i < strindex + mn; i++)
document.write(a[i] + " ");
}
// Driver program
let A=[1, 2, 2, 2, 1];
let n = A.length;
smallestSubsegment(A, n);
// This code is contributed by unknown2108
</script>
Output:
2 2 2
Time Complexity: O(n)
As we are doing linear operations on the given array.
Auxiliary Space: O(n)
The extra space is used to store the elements in the map, which in worst case can go upto O(n) when all elements are distinct.
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