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

  • Difficulty Level : Hard
  • Last Updated : 20 May, 2021
Geek Week

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:
 

  1. Consider all the 0’s in the array as ‘-1’.
  2. Initialize sum = 0 and maxLen = 0.
  3. Create a hash table having (sum, index) tuples.
  4. For i = 0 to n-1, perform the following steps:
    1. If arr[i] is ‘0’ accumulate ‘-1’ to sum else accumulate ‘1’ to sum.
    2. If sum == 1, update maxLen = i+1.
    3. 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.
    4. 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).
  5. Return maxLen.

 

C++




// 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's
int 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 above
int main()
{
    int arr[] = { 0, 1, 1, 0, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Length = "
         << lenOfLongSubarr(arr, n);
    return 0;
}

Java




// Java implementation to find the length of
// longest subarray having count of 1's one
// more than count of 0's
import java.util.*;
 
class GFG
{
 
// function to find the length of longest
// subarray having count of 1's one more
// than count of 0's
static 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 Code
public 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

Python3




# 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's
def 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 code
if __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#




// C# implementation to find the length of
// longest subarray having count of 1's one
// more than count of 0's
using 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's
static 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 Code
public 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

Javascript




<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's
function 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 above
var 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.
 

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