Longest substring with count of 1s more than 0s
Last Updated :
02 May, 2025
Given a binary string find the longest substring which contains 1’s more than 0’s.
Examples:
Input : 1010
Output : 3
Substring 101 has 1 occurring more number of times than 0.
Input : 101100
Output : 5
Substring 10110 has 1 occurring more number of times than 0.
A simple solution is to one by one consider all the substrings and check if that substring has a count of 1 more than 0. If the count is more than comparing its length with maximum length substring found till now. The time complexity of this solution is O(n^2).
An efficient solution is to use hashing. The idea is to find the sum of string traversed until now. Add 1 to the result if the current character is '1' else subtract 1. Now the problem reduces to finding the largest subarray having a sum greater than zero. To find the largest subarray having a sum greater than zero, we check the value of the sum. If sum is greater than zero, then the largest subarray with a sum greater than zero is arr[0..i].
If the sum is less than zero, then find the size of subarray arr[j+1..i], where j is index up to which sum of subarray arr[0..j] is sum -1 and j < i and compare that size with largest subarray size found so far. To find index j, store values of sum for arr[0..j] in hash table for all 0 <= j <= i. There might be a possibility that a given value of sum repeats. In that case store only first index for which that sum is obtained as it is required to get the length of largest subarray and that is obtained from first index occurrence.
Steps to solve this problem:
1. Declare n=bin.length ,i,sum=0.
2. Declare unordered map prevSum of key and value integer type.
3. Declare variables maxlen=0,currlen.
4. Iterate through i=0 till n:
*Check if bin[i]=1 than sum++ else sum--.
*Check if sum is greater than zero than maxlen=i+1.
*Else check if sum is smaller than equal to zero and sum-1 is present in prevSum than current=i-prevSum[sum-1] and maxlen=max(maxlen,currlen).
*Check if sum is not present in prevSum than prevSum[sum]=i.
5. Return maxlen.
Below is the implementation of above approach:
C++
// CPP program to find largest substring
// having count of 1s more than count
// count of 0s.
#include <bits/stdc++.h>
using namespace std;
// Function to find longest substring
// having count of 1s more than count
// of 0s.
int findLongestSub(string bin)
{
int n = bin.length(), i;
// To store sum.
int sum = 0;
// To store first occurrence of each
// sum value.
unordered_map<int, int> prevSum;
// To store maximum length.
int maxlen = 0;
// To store current substring length.
int currlen;
for (i = 0; i < n; i++) {
// Add 1 if current character is 1
// else subtract 1.
if (bin[i] == '1')
sum++;
else
sum--;
// If sum is positive, then maximum
// length substring is bin[0..i]
if (sum > 0) {
maxlen = i + 1;
}
// If sum is negative, then maximum
// length substring is bin[j+1..i], where
// sum of substring bin[0..j] is sum-1.
else if (sum <= 0) {
if (prevSum.find(sum - 1) != prevSum.end()) {
currlen = i - prevSum[sum - 1];
maxlen = max(maxlen, currlen);
}
}
// Make entry for this sum value in hash
// table if this value is not present.
if (prevSum.find(sum) == prevSum.end())
prevSum[sum] = i;
}
return maxlen;
}
// Driver code
int main()
{
string bin = "1010";
cout << findLongestSub(bin);
return 0;
}
// Java program to find largest substring
// having count of 1s more than count
// count of 0s.
import java.util.HashMap;
class GFG
{
// Function to find longest substring
// having count of 1s more than count
// of 0s.
static int findLongestSub(String bin)
{
int n = bin.length(), i;
// To store sum.
int sum = 0;
// To store first occurrence of each
// sum value.
HashMap<Integer,
Integer> prevSum = new HashMap<>();
// To store maximum length.
int maxlen = 0;
// To store current substring length.
int currlen;
for (i = 0; i < n; i++)
{
// Add 1 if current character is 1
// else subtract 1.
if (bin.charAt(i) == '1')
sum++;
else
sum--;
// If sum is positive, then maximum
// length substring is bin[0..i]
if (sum > 0)
{
maxlen = i + 1;
}
// If sum is negative, then maximum
// length substring is bin[j+1..i], where
// sum of substring bin[0..j] is sum-1.
else if (sum <= 0)
{
if (prevSum.containsKey(sum - 1))
{
currlen = i - (prevSum.get(sum - 1) == null ? 1 :
prevSum.get(sum - 1));
maxlen = Math.max(maxlen, currlen);
}
}
// Make entry for this sum value in hash
// table if this value is not present.
if (!prevSum.containsKey(sum))
prevSum.put(sum, i);
}
return maxlen;
}
// Driver code
public static void main(String[] args)
{
String bin = "1010";
System.out.println(findLongestSub(bin));
}
}
// This code is contributed by
// sanjeev2552
# Python 3 program to find largest substring
# having count of 1s more than count
# count of 0s.
# Function to find longest substring
# having count of 1s more than count
# of 0s.
def findLongestSub(bin1):
n = len(bin1)
# To store sum.
sum = 0
# To store first occurrence of each
# sum value.
prevSum = {i:0 for i in range(n)}
# To store maximum length.
maxlen = 0
# To store current substring length.
for i in range(n):
# Add 1 if current character is 1
# else subtract 1.
if (bin1[i] == '1'):
sum += 1
else:
sum -= 1
# If sum is positive, then maximum
# length substring is bin1[0..i]
if (sum > 0):
maxlen = i + 1
# If sum is negative, then maximum
# length substring is bin1[j+1..i], where
# sum of substring bin1[0..j] is sum-1.
elif (sum <= 0):
if ((sum - 1) in prevSum):
currlen = i - prevSum[sum - 1]
maxlen = max(maxlen, currlen)
# Make entry for this sum value in hash
# table if this value is not present.
if ((sum) not in prevSum):
prevSum[sum] = i
return maxlen
# Driver code
if __name__ == '__main__':
bin1 = "1010"
print(findLongestSub(bin1))
# This code is contributed by
# Surendra_Gangwar
// C# program to find largest substring
// having count of 1s more than count
// count of 0s.
using System;
using System.Collections.Generic;
class GFG
{
// Function to find longest substring
// having count of 1s more than count
// of 0s.
static int findLongestSub(String bin)
{
int n = bin.Length, i;
// To store sum.
int sum = 0;
// To store first occurrence of each
// sum value.
Dictionary<int,
int> prevSum = new Dictionary<int,
int>();
// To store maximum length.
int maxlen = 0;
// To store current substring length.
int currlen;
for (i = 0; i < n; i++)
{
// Add 1 if current character is 1
// else subtract 1.
if (bin[i] == '1')
sum++;
else
sum--;
// If sum is positive, then maximum
// length substring is bin[0..i]
if (sum > 0)
{
maxlen = i + 1;
}
// If sum is negative, then maximum
// length substring is bin[j+1..i], where
// sum of substring bin[0..j] is sum-1.
else if (sum <= 0)
{
if (prevSum.ContainsKey(sum - 1))
{
currlen = i - (prevSum[sum - 1] == 0 ? 1 :
prevSum[sum - 1]);
maxlen = Math.Max(maxlen, currlen);
}
}
// Make entry for this sum value in hash
// table if this value is not present.
if (!prevSum.ContainsKey(sum))
prevSum.Add(sum, i);
}
return maxlen;
}
// Driver code
public static void Main(String[] args)
{
String bin = "1010";
Console.WriteLine(findLongestSub(bin));
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program to find largest substring
// having count of 1s more than count
// count of 0s.
// Function to find longest substring
// having count of 1s more than count
// of 0s.
function findLongestSub(bin)
{
let n = bin.length, i;
// To store sum.
let sum = 0;
// To store first occurrence of each
// sum value.
let prevSum = new Map();
// To store maximum length.
let maxlen = 0;
// To store current substring length.
let currlen;
for (i = 0; i < n; i++)
{
// Add 1 if current character is 1
// else subtract 1.
if (bin[i] == '1')
sum++;
else
sum--;
// If sum is positive, then maximum
// length substring is bin[0..i]
if (sum > 0)
{
maxlen = i + 1;
}
// If sum is negative, then maximum
// length substring is bin[j+1..i], where
// sum of substring bin[0..j] is sum-1.
else if (sum <= 0)
{
if (prevSum.has(sum - 1))
{
currlen = i - (prevSum.get(sum - 1) == null ? 1 :
prevSum.get(sum - 1));
maxlen = Math.max(maxlen, currlen);
}
}
// Make entry for this sum value in hash
// table if this value is not present.
if (!prevSum.has(sum))
prevSum.set(sum, i);
}
return maxlen;
}
// Driver code
let bin = "1010";
document.write(findLongestSub(bin));
// This code is contributed by rag2127
</script>
complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)
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