Longest subarray having count of 1s one more than count of 0s - GeeksforGeeks

20 May, 2021
Given an array of size n containing 0’s and 1’s only. The problem is to find the length of the longest subarray having count of 1’s one more than count of 0’s. Examples:  
Input : arr = {0, 1, 1, 0, 0, 1}
Output : 5
From index 1 to 5.

Input : arr[] = {1, 0, 0, 1, 0}
Output : 1
 
Approach: Following are the steps: 
Consider all the 0’s in the array as ‘-1’.Initialize sum = 0 and maxLen = 0.Create a hash table having (sum, index) tuples.For i = 0 to n-1, perform the following steps:If arr[i] is ‘0’ accumulate ‘-1’ to sum else accumulate ‘1’ to sum.If sum == 1, update maxLen = i+1.Else check whether sum is present in the hash table or not. If not present, then add it to the hash table as (sum, i) pair.Check if (sum-1) is present in the hash table or not. if present, then obtain index of (sum-1) from the hash table as index. Now check if maxLen is less than (i-index), then update maxLen = (i-index).Return maxLen.
Consider all the 0’s in the array as ‘-1’.
Initialize sum = 0 and maxLen = 0.
Create a hash table having (sum, index) tuples.
For i = 0 to n-1, perform the following steps:If arr[i] is ‘0’ accumulate ‘-1’ to sum else accumulate ‘1’ to sum.If sum == 1, update maxLen = i+1.Else check whether sum is present in the hash table or not. If not present, then add it to the hash table as (sum, i) pair.Check if (sum-1) is present in the hash table or not. if present, then obtain index of (sum-1) from the hash table as index. Now check if maxLen is less than (i-index), then update maxLen = (i-index).
If arr[i] is ‘0’ accumulate ‘-1’ to sum else accumulate ‘1’ to sum.If sum == 1, update maxLen = i+1.Else check whether sum is present in the hash table or not. If not present, then add it to the hash table as (sum, i) pair.Check if (sum-1) is present in the hash table or not. if present, then obtain index of (sum-1) from the hash table as index. Now check if maxLen is less than (i-index), then update maxLen = (i-index).
If arr[i] is ‘0’ accumulate ‘-1’ to sum else accumulate ‘1’ to sum.
If sum == 1, update maxLen = i+1.
Else check whether sum is present in the hash table or not. If not present, then add it to the hash table as (sum, i) pair.
Check if (sum-1) is present in the hash table or not. if present, then obtain index of (sum-1) from the hash table as index. Now check if maxLen is less than (i-index), then update maxLen = (i-index).
Return maxLen.
 
C++
Java
Python3
C#
Javascript
// C++ implementation to find the length of// longest subarray having count of 1's one// more than count of 0's#include <bits/stdc++.h>using namespace std; // function to find the length of longest// subarray having count of 1's one more// than count of 0'sint lenOfLongSubarr(int arr[], int n){    // unordered_map 'um' implemented as    // hash table    unordered_map<int, int> um;    int sum = 0, maxLen = 0;     // traverse the given array    for (int i = 0; i < n; i++) {         // consider '0' as '-1'        sum += arr[i] == 0 ? -1 : 1;         // when subarray starts form index '0'        if (sum == 1)            maxLen = i + 1;         // make an entry for 'sum' if it is        // not present in 'um'        else if (um.find(sum) == um.end())            um[sum] = i;         // check if 'sum-1' is present in 'um'        // or not        if (um.find(sum - 1) != um.end()) {             // update maxLength            if (maxLen < (i - um[sum - 1]))                maxLen = i - um[sum - 1];        }    }     // required maximum length    return maxLen;} // Driver program to test aboveint main(){    int arr[] = { 0, 1, 1, 0, 0, 1 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << "Length = "         << lenOfLongSubarr(arr, n);    return 0;}
// Java implementation to find the length of// longest subarray having count of 1's one// more than count of 0'simport java.util.*; class GFG{ // function to find the length of longest// subarray having count of 1's one more// than count of 0'sstatic int lenOfLongSubarr(int arr[], int n){    // unordered_map 'um' implemented as    // hash table    HashMap<Integer,            Integer> um = new HashMap<Integer,                                      Integer>();    int sum = 0, maxLen = 0;     // traverse the given array    for (int i = 0; i < n; i++)    {         // consider '0' as '-1'        sum += arr[i] == 0 ? -1 : 1;         // when subarray starts form index '0'        if (sum == 1)            maxLen = i + 1;         // make an entry for 'sum' if it is        // not present in 'um'        else if (!um.containsKey(sum))            um. put(sum, i);         // check if 'sum-1' is present in 'um'        // or not        if (um.containsKey(sum - 1))        {             // update maxLength            if (maxLen < (i - um.get(sum - 1)))                maxLen = i - um.get(sum - 1);        }    }     // required maximum length    return maxLen;} // Driver Codepublic static void main(String[] args){    int arr[] = { 0, 1, 1, 0, 0, 1 };    int n = arr.length;    System.out.println("Length = " +               lenOfLongSubarr(arr, n));}} // This code is contributed by Princi Singh
# Python 3 implementation to find the length of# longest subarray having count of 1's one# more than count of 0's # function to find the length of longest# subarray having count of 1's one more# than count of 0'sdef lenOfLongSubarr(arr, n):         # unordered_map 'um' implemented as    # hash table    um = {i:0 for i in range(10)}    sum = 0    maxLen = 0     # traverse the given array    for i in range(n):                 # consider '0' as '-1'        if arr[i] == 0:            sum += -1        else:            sum += 1         # when subarray starts form index '0'        if (sum == 1):            maxLen = i + 1         # make an entry for 'sum' if it is        # not present in 'um'        elif (sum not in um):            um[sum] = i         # check if 'sum-1' is present in 'um'        # or not        if ((sum - 1) in um):            # update maxLength            if (maxLen < (i - um[sum - 1])):                maxLen = i - um[sum - 1]     # required maximum length    return maxLen # Driver codeif __name__ == '__main__':    arr = [0, 1, 1, 0, 0, 1]    n = len(arr)    print("Length =",lenOfLongSubarr(arr, n))     # This code is contributed by# Surendra_Gangwar
// C# implementation to find the length of// longest subarray having count of 1's one// more than count of 0'susing System;using System.Collections.Generic;     class GFG{ // function to find the length of longest// subarray having count of 1's one more// than count of 0'sstatic int lenOfLongSubarr(int []arr, int n){    // unordered_map 'um' implemented as    // hash table    Dictionary<int,               int> um = new Dictionary<int,                                        int>();    int sum = 0, maxLen = 0;     // traverse the given array    for (int i = 0; i < n; i++)    {         // consider '0' as '-1'        sum += arr[i] == 0 ? -1 : 1;         // when subarray starts form index '0'        if (sum == 1)            maxLen = i + 1;         // make an entry for 'sum' if it is        // not present in 'um'        else if (!um.ContainsKey(sum))            um.Add(sum, i);         // check if 'sum-1' is present in 'um'        // or not        if (um.ContainsKey(sum - 1))        {             // update maxLength            if (maxLen < (i - um[sum - 1]))                maxLen = i - um[sum - 1];        }    }     // required maximum length    return maxLen;} // Driver Codepublic static void Main(String[] args){    int []arr = { 0, 1, 1, 0, 0, 1 };    int n = arr.Length;    Console.WriteLine("Length = " +            lenOfLongSubarr(arr, n));}} // This code is contributed by Rajput-Ji
<script>  // Javascript implementation to find the length of// longest subarray having count of 1's one// more than count of 0's // function to find the length of longest// subarray having count of 1's one more// than count of 0'sfunction lenOfLongSubarr(arr, n){    // unordered_map 'um' implemented as    // hash table    var um = new Map();    var sum = 0, maxLen = 0;     // traverse the given array    for (var i = 0; i < n; i++) {         // consider '0' as '-1'        sum += arr[i] == 0 ? -1 : 1;         // when subarray starts form index '0'        if (sum == 1)            maxLen = i + 1;         // make an entry for 'sum' if it is        // not present in 'um'        else if (!um.has(sum))            um.set(sum, i);         // check if 'sum-1' is present in 'um'        // or not        if (um.has(sum - 1)) {             // update maxLength            if (maxLen < (i - um.get(sum - 1)))                maxLen = i - um.get(sum - 1);        }    }     // required maximum length    return maxLen;} // Driver program to test abovevar arr = [0, 1, 1, 0, 0, 1];var n = arr.length;document.write( "Length = "      + lenOfLongSubarr(arr, n)); // This code is contributed by itsok.</script>
Output:  
Length = 5
Time Complexity: O(n) Auxiliary Space: O(n)This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
SURENDRA_GANGWAR
princi singh
Rajput-Ji
itsok
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