Largest subarray with equal number of 0s and 1s

Given an array containing only 0s and 1s, find the largest subarray which contain equal no of 0s and 1s. Expected time complexity is O(n).

Examples:

Input: arr[] = {1, 0, 1, 1, 1, 0, 0}
Output: 1 to 6 (Starting and Ending indexes of output subarray)

Input: arr[] = {1, 1, 1, 1}
Output: No such subarray

Input: arr[] = {0, 0, 1, 1, 0}
Output: 0 to 3 Or 1 to 4

Method 1 (Simple)
A simple method is to use two nested loops. The outer loop picks a starting point i. The inner loop considers all subarrays starting from i. If size of a subarray is greater than maximum size so far, then update the maximum size.
In the below code, 0s are considered as -1 and sum of all values from i to j is calculated. If sum becomes 0, then size of this subarray is compared with largest size so far.

C++

// A simple C++ program to find the largest subarray  
// with equal number of 0s and 1s  
#include<bits/stdc++.h>  
  
using namespace std;  
  
// This function Prints the starting and ending  
// indexes of the largest subarray with equal  
// number of 0s and 1s. Also returns the size  
// of such subarray.  
  
int findSubArray(int arr[], int n)  
{  
    int sum = 0;  
    int maxsize = -1, startindex;  
  
    // Pick a starting point as i  
    for (int i = 0; i < n-1; i++)  
    {  
        sum = (arr[i] == 0)? -1 : 1;  
  
        // Consider all subarrays starting from i  
        for (int j = i+1; j < n; j++)  
        {  
            (arr[j] == 0)? (sum += -1): (sum += 1);  
  
            // If this is a 0 sum subarray, then  
            // compare it with maximum size subarray  
            // calculated so far  
            if (sum == 0 && maxsize < j-i+1)  
            {  
                maxsize = j - i + 1;  
                startindex = i;  
            }  
        }  
    }  
    if (maxsize == -1)  
        cout << "No such subarray";  
    else
        cout << startindex << " to " << startindex + maxsize - 1;  
  
    return maxsize;  
}  
  
/* Driver code*/
int main()  
{  
    int arr[] = {1, 0, 0, 1, 0, 1, 1};  
    int size = sizeof(arr)/sizeof(arr[0]);  
  
    findSubArray(arr, size);  
    return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

// A simple program to find the largest subarray  
// with equal number of 0s and 1s 
  
#include <stdio.h> 
  
// This function Prints the starting and ending 
// indexes of the largest subarray with equal  
// number of 0s and 1s. Also returns the size  
// of such subarray. 
  
int findSubArray(int arr[], int n) 
{ 
    int sum = 0; 
    int maxsize = -1, startindex; 
  
    // Pick a starting point as i 
  
    for (int i = 0; i < n-1; i++) 
    { 
        sum = (arr[i] == 0)? -1 : 1; 
  
        // Consider all subarrays starting from i 
  
        for (int j = i+1; j < n; j++) 
        { 
            (arr[j] == 0)? (sum += -1): (sum += 1); 
  
            // If this is a 0 sum subarray, then  
            // compare it with maximum size subarray 
            // calculated so far 
  
            if (sum == 0 && maxsize < j-i+1) 
            { 
                maxsize = j - i + 1; 
                startindex = i; 
            } 
        } 
    } 
    if (maxsize == -1) 
        printf("No such subarray"); 
    else
        printf("%d to %d", startindex, startindex+maxsize-1); 
  
    return maxsize; 
} 
  
/* Driver program to test above functions*/
  
int main() 
{ 
    int arr[] =  {1, 0, 0, 1, 0, 1, 1}; 
    int size = sizeof(arr)/sizeof(arr[0]); 
  
    findSubArray(arr, size); 
    return 0; 
} 

Java

class LargestSubArray  
{ 
  
    // This function Prints the starting and ending 
    // indexes of the largest subarray with equal   
    // number of 0s and 1s. Also returns the size  
    // of such subarray. 
  
    int findSubArray(int arr[], int n)  
    { 
        int sum = 0; 
        int maxsize = -1, startindex = 0; 
        int endindex = 0; 
  
        // Pick a starting point as i 
  
        for (int i = 0; i < n - 1; i++)  
        { 
            sum = (arr[i] == 0) ? -1 : 1; 
  
            // Consider all subarrays starting from i 
  
            for (int j = i + 1; j < n; j++)  
            { 
                if(arr[j] == 0)   
                    sum += -1;  
                else
                    sum += 1; 
  
                // If this is a 0 sum subarray, then  
                // compare it with maximum size subarray 
                // calculated so far 
  
                if (sum == 0 && maxsize < j - i + 1)  
                { 
                    maxsize = j - i + 1; 
                    startindex = i; 
                } 
            } 
        } 
        endindex = startindex+maxsize-1; 
        if (maxsize == -1) 
            System.out.println("No such subarray"); 
        else
            System.out.println(startindex+" to "+endindex); 
  
        return maxsize; 
    } 
  
    /* Driver program to test the above functions */
  
    public static void main(String[] args)  
    { 
        LargestSubArray sub; 
        sub = new LargestSubArray(); 
        int arr[] = {1, 0, 0, 1, 0, 1, 1}; 
        int size = arr.length; 
  
        sub.findSubArray(arr, size); 
    } 
} 

Python3

# A simple program to find the largest subarray 
# with equal number of 0s and 1s 
  
# This function Prints the starting and ending 
# indexes of the largest subarray with equal  
# number of 0s and 1s. Also returns the size  
# of such subarray. 
def findSubArray(arr, n): 
  
    sum = 0
    maxsize = -1
  
    # Pick a starting point as i 
  
    for i in range(0, n-1): 
      
        sum = -1 if(arr[i] == 0) else 1
  
        # Consider all subarrays starting from i 
  
        for j in range(i + 1, n): 
          
            sum = sum + (-1) if (arr[j] == 0) else sum + 1
  
            # If this is a 0 sum subarray, then  
            # compare it with maximum size subarray 
            # calculated so far 
  
            if (sum == 0 and maxsize < j-i + 1): 
                  
                maxsize = j - i + 1
                startindex = i 
              
          
      
    if (maxsize == -1): 
        print("No such subarray"); 
    else: 
        print(startindex, "to", startindex + maxsize-1); 
  
    return maxsize 
  
# Driver program to test above functions 
arr = [1, 0, 0, 1, 0, 1, 1] 
size = len(arr) 
findSubArray(arr, size) 
  
# This code is contributed by Smitha Dinesh Semwal 

C#

// A simple program to find the largest subarray  
// with equal number of 0s and 1s 
using System; 
  
class GFG 
{ 
              
    // This function Prints the starting and ending 
    // indexes of the largest subarray with equal  
    // number of 0s and 1s. Also returns the size  
    // of such subarray. 
  
    static int findSubArray(int []arr, int n)  
    { 
        int sum = 0; 
        int maxsize = -1, startindex = 0; 
        int endindex = 0; 
  
        // Pick a starting point as i 
        for (int i = 0; i < n - 1; i++)  
        { 
            sum = (arr[i] == 0) ? -1 : 1; 
  
            // Consider all subarrays starting from i 
  
            for (int j = i + 1; j < n; j++)  
            { 
                if(arr[j] == 0)  
                    sum += -1;  
                else
                    sum += 1; 
  
                // If this is a 0 sum subarray, then  
                // compare it with maximum size subarray 
                // calculated so far 
  
                if (sum == 0 && maxsize < j - i + 1)  
                { 
                    maxsize = j - i + 1; 
                    startindex = i; 
                } 
            } 
        } 
        endindex = startindex+maxsize-1; 
        if (maxsize == -1) 
            Console.WriteLine("No such subarray"); 
        else
            Console.WriteLine(startindex+" to "+endindex); 
  
        return maxsize; 
    } 
  
    // Driver program  
    public static void Main()  
    { 
          
        int []arr = {1, 0, 0, 1, 0, 1, 1}; 
        int size = arr.Length; 
        findSubArray(arr, size); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php  
// A simple program to find the  
// largest subarray with equal  
// number of 0s and 1s 
  
// This function Prints the starting  
// and ending indexes of the largest  
// subarray with equal number of 0s  
// and 1s. Also returns the size of  
// such subarray. 
function findSubArray(&$arr, $n) 
{ 
    $sum = 0; 
    $maxsize = -1; 
  
    // Pick a starting point as i 
  
    for ($i = 0; $i < $n - 1; $i++) 
    { 
        $sum = ($arr[$i] == 0) ? -1 : 1; 
  
        // Consider all subarrays 
        // starting from i 
        for ($j = $i + 1; $j < $n; $j++) 
        { 
            ($arr[$j] == 0) ?  
               ($sum += -1) : ($sum += 1); 
  
            // If this is a 0 sum subarray,  
            // then compare it with maximum  
            // size subarray calculated so far 
  
            if ($sum == 0 && $maxsize < $j - $i + 1) 
            { 
                $maxsize = $j - $i + 1; 
                $startindex = $i; 
            } 
        } 
    } 
    if ($maxsize == -1) 
        echo "No such subarray"; 
    else
        echo $startindex. " to " . 
            ($startindex + $maxsize - 1); 
  
    return $maxsize; 
} 
  
// Driver Code 
$arr = array(1, 0, 0, 1, 0, 1, 1); 
$size = sizeof($arr); 
  
findSubArray($arr, $size); 
  
// This coed is contributed  
// by ChitraNayal 
?> 

Output:

 0 to 5

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

Method 2 (Tricky)
Following is a solution that uses O(n) extra space and solves the problem in O(n) time complexity.
Let input array be arr[] of size n and maxsize be the size of output subarray.
1) Consider all 0 values as -1. The problem now reduces to find out the maximum length subarray with sum = 0.
2) Create a temporary array sumleft[] of size n. Store the sum of all elements from arr[0] to arr[i] in sumleft[i]. This can be done in O(n) time.
3) There are two cases, the output subarray may start from 0th index or may start from some other index. We will return the max of the values obtained by two cases.
4) To find the maximum length subarray starting from 0th index, scan the sumleft[] and find the maximum i where sumleft[i] = 0.
5) Now, we need to find the subarray where subarray sum is 0 and start index is not 0. This problem is equivalent to finding two indexes i & j in sumleft[] such that sumleft[i] = sumleft[j] and j-i is maximum. To solve this, we can create a hash table with size = max-min+1 where min is the minimum value in the sumleft[] and max is the maximum value in the sumleft[]. The idea is to hash the leftmost occurrences of all different values in sumleft[]. The size of hash is chosen as max-min+1 because there can be these many different possible values in sumleft[]. Initialize all values in hash as -1
6) To fill and use hash[], traverse sumleft[] from 0 to n-1. If a value is not present in hash[], then store its index in hash. If the value is present, then calculate the difference of current index of sumleft[] and previously stored value in hash[]. If this difference is more than maxsize, then update the maxsize.
7) To handle corner cases (all 1s and all 0s), we initialize maxsize as -1. If the maxsize remains -1, then print there is no such subarray.

C

// A O(n) program to find the largest subarray  
// with equal number of 0s and 1s 
  
#include <stdio.h> 
#include <stdlib.h> 
   
// A utility function to get maximum of two  
// integers 
  
int max(int a, int b) { return a>b? a: b; } 
   
// This function Prints the starting and ending  
// indexes of the largest subarray with equal 
// number of 0s and 1s. Also returns the size 
// of such subarray. 
  
int findSubArray(int arr[], int n) 
{ 
    // variables to store result values 
  
    int maxsize = -1, startindex;   
   
    // Create an auxiliary array sunmleft[].  
    // sumleft[i] will be sum of array  
    // elements from arr[0] to arr[i] 
  
    int sumleft[n]; 
  
    // For min and max values in sumleft[] 
  
    int min, max;  
    int i; 
   
    // Fill sumleft array and get min and max  
    // values in it.  Consider 0 values in arr[] 
    // as -1 
  
    sumleft[0] = ((arr[0] == 0)? -1: 1); 
    min = arr[0]; max = arr[0]; 
    for (i=1; i<n; i++) 
    {       
        sumleft[i] = sumleft[i-1] + ((arr[i] == 0)?  
                     -1: 1); 
        if (sumleft[i] < min) 
            min = sumleft[i]; 
        if (sumleft[i] > max) 
            max = sumleft[i]; 
    } 
   
    // Now calculate the max value of j - i such  
    // that sumleft[i] = sumleft[j]. The idea is  
    // to create a hash table to store indexes of all 
    // visited values.    
    // If you see a value again, that it is a case of  
    // sumleft[i] = sumleft[j]. Check if this j-i is  
    // more than maxsize.  
    // The optimum size of hash will be max-min+1 as  
    // these many different values of sumleft[i] are 
    // possible. Since we use optimum size, we need  
    // to shift all values in sumleft[] by min before  
    // using them as an index in hash[]. 
  
    int hash[max-min+1]; 
   
    // Initialize hash table 
  
    for (i=0; i<max-min+1; i++) 
        hash[i] = -1; 
   
    for (i=0; i<n; i++) 
    { 
        // Case 1: when the subarray starts from  
        //         index 0 
  
        if (sumleft[i] == 0) 
        { 
           maxsize = i+1; 
           startindex = 0; 
        } 
   
        // Case 2: fill hash table value. If already 
        //         filled, then use it 
  
        if (hash[sumleft[i]-min] == -1) 
            hash[sumleft[i]-min] = i; 
        else
        { 
            if ((i - hash[sumleft[i]-min]) > maxsize) 
            { 
                maxsize = i - hash[sumleft[i]-min]; 
                startindex = hash[sumleft[i]-min] + 1; 
            } 
        } 
    } 
    if (maxsize == -1) 
        printf("No such subarray"); 
    else
        printf("%d to %d", startindex, startindex+maxsize-1); 
   
    return maxsize; 
} 
   
/* Driver program to test above functions */
int main() 
{ 
    int arr[] =  {1, 0, 0, 1, 0, 1, 1}; 
    int size = sizeof(arr)/sizeof(arr[0]); 
   
    findSubArray(arr, size); 
    return 0; 
} 

C++/STL

// C++ program to find largest subarray with equal number of 
// 0's and 1's. 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns largest subarray with equal number of 0s and 1s 
  
int maxLen(int arr[], int n) 
{ 
    // Creates an empty hashMap hM 
  
    unordered_map<int, int> hM; 
  
    int sum = 0;     // Initialize sum of elements 
    int max_len = 0; // Initialize result 
    int ending_index = -1; 
  
    for (int i = 0; i < n; i++) 
        arr[i] = (arr[i] == 0)? -1: 1; 
  
    // Traverse through the given array 
  
    for (int i = 0; i < n; i++) 
    { 
        // Add current element to sum 
  
        sum += arr[i]; 
  
        // To handle sum=0 at last index 
  
        if (sum == 0) 
        { 
            max_len = i + 1; 
            ending_index = i; 
        } 
  
        // If this sum is seen before, then update max_len 
        // if required 
  
        if (hM.find(sum + n) != hM.end()) 
        { 
            if (max_len < i - hM[sum + n]) 
            { 
                max_len = i - hM[sum + n]; 
                ending_index = i; 
            } 
        } 
        else // Else put this sum in hash table 
            hM[sum + n] = i; 
    } 
  
    for (int i = 0; i < n; i++) 
        arr[i] = (arr[i] == -1)? 0: 1; 
  
    printf("%d to %d
", ending_index-max_len+1, ending_index); 
  
    return max_len; 
} 
  
// Driver method 
  
int main() 
{ 
    int arr[] = {1, 0, 0, 1, 0, 1, 1}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    maxLen(arr, n); 
    return 0; 
} 
  
// This code is contributed by Aditya Goel 

Java

import java.util.HashMap; 
  
class LargestSubArray1  
{ 
  
    // Returns largest subarray with equal number of 0s and 1s 
   
    int maxLen(int arr[], int n)  
    { 
        // Creates an empty hashMap hM 
   
        HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>(); 
  
        int sum = 0;     // Initialize sum of elements 
        int max_len = 0; // Initialize result 
        int ending_index = -1; 
        int start_index = 0; 
  
        for (int i = 0; i < n; i++)  
        { 
            arr[i] = (arr[i] == 0) ? -1 : 1; 
        } 
  
        // Traverse through the given array 
   
        for (int i = 0; i < n; i++)  
        { 
            // Add current element to sum 
   
            sum += arr[i]; 
  
            // To handle sum=0 at last index 
   
            if (sum == 0)  
            { 
                max_len = i + 1; 
                ending_index = i; 
            } 
  
            // If this sum is seen before, then update max_len 
            // if required 
   
            if (hM.containsKey(sum + n))  
            { 
                if (max_len < i - hM.get(sum + n))  
                { 
                    max_len = i - hM.get(sum + n); 
                    ending_index = i; 
                } 
            }  
            else // Else put this sum in hash table 
                hM.put(sum + n, i); 
        } 
  
        for (int i = 0; i < n; i++)  
        { 
            arr[i] = (arr[i] == -1) ? 0 : 1; 
        } 
  
        int end = ending_index - max_len + 1; 
        System.out.println(end + " to " + ending_index); 
  
        return max_len; 
    } 
  
    /* Driver program to test the above functions */
   
    public static void main(String[] args)  
    { 
        LargestSubArray1 sub = new LargestSubArray1(); 
        int arr[] = {1, 0, 0, 1, 0, 1, 1}; 
        int n = arr.length; 
  
        sub.maxLen(arr, n); 
    } 
} 
  
// This code has been by Mayank Jaiswal(mayank_24) 

Python3

# Python 3 program to find largest  
# subarray with equal number of 
# 0's and 1's. 
  
# Returns largest subarray with  
# equal number of 0s and 1s 
def maxLen(arr, n): 
  
    # NOTE: Dictonary in python in  
    # implemented as Hash Maps.  
    # Create an empty hash map (dictionary)  
    hash_map = {} ; 
    curr_sum = 0;  
    max_len = 0;  
    ending_index = -1; 
  
    for i in range (0, n): 
        if(arr[i] == 0): 
            arr[i] = -1; 
        else: 
            arr[i] = 1; 
  
    # Traverse through the given array 
    for i in range (0, n): 
      
        # Add current element to sum 
        curr_sum = curr_sum + arr[i]; 
  
        # To handle sum=0 at last index 
        if (curr_sum == 0): 
            max_len = i + 1; 
            ending_index = i; 
  
        # If this sum is seen before, 
        # then update max_len if required 
        if (curr_sum + n) in hash_map:  
            max_len = max(max_len, i -
                          hash_map[curr_sum + n])  
        else:  
  
            # else put this sum in dictionary  
            hash_map[curr_sum] = i ; 
          
    for i in range (0, n): 
        if(arr[i] == -1): 
            arr[i] = 0; 
        else: 
            arr[i] = 1; 
              
    print (ending_index - max_len + 1, 
                            end =" "); 
    print ("to", end = " "); 
    print (ending_index); 
  
    return max_len; 
  
# Driver Code 
arr = [1, 0, 0, 1, 0, 1, 1]; 
n = len(arr) ; 
  
maxLen(arr, n); 
      
# This code is contributed 
# by Shivi_Aggarwal 

C#

// C# program to find the largest subarray  
// with equal number of 0s and 1s 
using System; 
using System.Collections.Generic; 
  
class LargestSubArray1 { 
  
    // Returns largest subarray with 
    // equal number of 0s and 1s  
    public virtual int maxLen(int[] arr, int n) { 
          
        // Creates an empty Dictionary hM  
        Dictionary < int, 
        int > hM = new Dictionary < int, 
        int > (); 
  
        int sum = 0; // Initialize sum of elements 
        int max_len = 0; // Initialize result 
        int ending_index = -1; 
        int start_index = 0; 
  
        for (int i = 0; i < n; i++) { 
            arr[i] = (arr[i] == 0) ? -1 : 1; 
        } 
  
        // Traverse through the given array  
        for (int i = 0; i < n; i++) { 
            // Add current element to sum  
            sum += arr[i]; 
  
            // To handle sum=0 at last index  
            if (sum == 0) { 
                max_len = i + 1; 
                ending_index = i; 
            } 
  
            // If this sum is seen before,  
            // then update max_len  
            // if required  
            if (hM.ContainsKey(sum + n)) { 
                if (max_len < i - hM[sum + n]) { 
                    max_len = i - hM[sum + n]; 
                    ending_index = i; 
                } 
            } 
              
            else // Else put this sum in hash table 
            { 
                hM[sum + n] = i; 
            } 
        } 
  
        for (int i = 0; i < n; i++) { 
            arr[i] = (arr[i] == -1) ? 0 : 1; 
        } 
  
        int end = ending_index - max_len + 1; 
        Console.WriteLine(end + " to " + ending_index); 
  
        return max_len; 
    } 
  
      
// Driver Code 
public static void Main(string[] args)  
{ 
      
    LargestSubArray1 sub = new LargestSubArray1(); 
    int[] arr = new int[]  
    { 
        1, 
        0, 
        0, 
        1, 
        0, 
        1, 
        1 
    }; 
      
    int n = arr.Length; 
    sub.maxLen(arr, n); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

 
0 to 5

Time Complexity: O(n)
Auxiliary Space: O(n)
Thanks to Aashish Barnwal for suggesting this solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Largest subarray with equal number of 0s and 1s

C++

C

Java

Python3

C#

PHP

C

C++/STL

Java

Python3

C#

tags:

leave a comment

0 Comments

Subscribe to Our Newsletter

Top Followed books