Open In App

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

Last Updated : 19 Oct, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

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.

Below is the implementation of the above approach:

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 from 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 = {}
    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)

using Brute Force:

Approach:

One way to solve this problem is to iterate through all possible subarrays and count the number of ones and zeros in each subarray. Then, we can check if the count of ones is one more than the count of zeros and keep track of the maximum subarray length that satisfies this condition.
 

Initialize the maximum subarray length to zero.
Iterate through all possible starting indices of the subarrays.
For each starting index, initialize the count of ones and zeros to zero.
Iterate through all possible ending indices of the subarrays.
For each ending index, update the counts of ones and zeros based on the value at the current index.
Check if the count of ones is one more than the count of zeros.
If the condition is satisfied, update the maximum subarray length.
Return the maximum subarray length.

C++




#include <bits/stdc++.h>
using namespace std;
 
// Added by ~Nikunj Sonigara
 
int longestSubarray(vector<int>& arr) {
    int n = arr.size();
    int maxLen = 0;
     
    for (int i = 0; i < n; i++) {
        int count0 = 0;
        int count1 = 0;
         
        for (int j = i; j < n; j++) {
            if (arr[j] == 0) {
                count0++;
            } else {
                count1++;
            }
             
            if (count1 == count0 + 1) {
                maxLen = max(maxLen, j - i + 1);
            }
        }
    }
     
    return maxLen;
}
 
int main() {
    vector<int> arr1 = {0, 1, 1, 0, 0, 1};
    cout << longestSubarray(arr1) << endl; // Output: 5
 
    vector<int> arr2 = {1, 0, 0, 1, 0};
    cout << longestSubarray(arr2) << endl; // Output: 1
 
    return 0;
}


Java




import java.util.*;
 
// Added by ~Nikunj Sonigara
 
public class Main {
    public static int longestSubarray(List<Integer> arr) {
        int n = arr.size();
        int maxLen = 0;
 
        for (int i = 0; i < n; i++) {
            int count0 = 0;
            int count1 = 0;
 
            for (int j = i; j < n; j++) {
                if (arr.get(j) == 0) {
                    count0++;
                } else {
                    count1++;
                }
 
                if (count1 == count0 + 1) {
                    maxLen = Math.max(maxLen, j - i + 1);
                }
            }
        }
 
        return maxLen;
    }
 
    public static void main(String[] args) {
        List<Integer> arr1 = new ArrayList<>();
        arr1.add(0);
        arr1.add(1);
        arr1.add(1);
        arr1.add(0);
        arr1.add(0);
        arr1.add(1);
        System.out.println(longestSubarray(arr1)); // Output: 5
 
        List<Integer> arr2 = new ArrayList<>();
        arr2.add(1);
        arr2.add(0);
        arr2.add(0);
        arr2.add(1);
        arr2.add(0);
        System.out.println(longestSubarray(arr2)); // Output: 1
    }
}


Python3




def longest_subarray(arr):
    n = len(arr)
    max_len = 0
    for i in range(n):
        count_0 = 0
        count_1 = 0
        for j in range(i, n):
            if arr[j] == 0:
                count_0 += 1
            else:
                count_1 += 1
            if count_1 == count_0 + 1:
                max_len = max(max_len, j - i + 1)
    return max_len
 
arr = [0, 1, 1, 0, 0, 1]
print(longest_subarray(arr))  # Output: 5
 
arr = [1, 0, 0, 1, 0]
print(longest_subarray(arr))  # Output: 1


C#




using System;
using System.Collections.Generic;
 
class Program {
    // Function to find the length of the longest subarray
    static int LongestSubarray(List<int> arr)
    {
        int n = arr.Count;
        int maxLen = 0;
 
        // Iterate through the array
        for (int i = 0; i < n; i++) {
            int count0 = 0;
            int count1 = 0;
 
            // Consider subarrays starting from index i
            for (int j = i; j < n; j++) {
                // Count the number of 0s and 1s in the
                // subarray
                if (arr[j] == 0) {
                    count0++;
                }
                else {
                    count1++;
                }
 
                // Check if the subarray has one more 1 than
                // 0
                if (count1 == count0 + 1) {
                    // Update the maximum length if the
                    // condition is met
                    maxLen = Math.Max(maxLen, j - i + 1);
                }
            }
        }
 
        return maxLen;
    }
 
    static void Main()
    {
        List<int> arr1 = new List<int>{ 0, 1, 1, 0, 0, 1 };
        Console.WriteLine(
            LongestSubarray(arr1)); // Output: 5
 
        List<int> arr2 = new List<int>{ 1, 0, 0, 1, 0 };
        Console.WriteLine(
            LongestSubarray(arr2)); // Output: 1
    }
}


Javascript




// Added by ~Nikunj Sonigara
 
function longestSubarray(arr) {
    const n = arr.length;
    let maxLen = 0;
 
    for (let i = 0; i < n; i++) {
        let count0 = 0;
        let count1 = 0;
 
        for (let j = i; j < n; j++) {
            if (arr[j] === 0) {
                count0++;
            } else {
                count1++;
            }
 
            if (count1 === count0 + 1) {
                maxLen = Math.max(maxLen, j - i + 1);
            }
        }
    }
 
    return maxLen;
}
 
const arr1 = [0, 1, 1, 0, 0, 1];
console.log(longestSubarray(arr1)); // Output: 5
 
const arr2 = [1, 0, 0, 1, 0];
console.log(longestSubarray(arr2)); // Output: 1


Output

5
1

Time Complexity: O(n^2)
Space Complexity: O(1)




Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads