# Maximum length of segments of 0’s and 1’s in C++

## Problem statement

Given a string comprising of ones and zeros. The task is to find the maximum length of the segments of string such that a number of 1 in each segment is greater than 0

## Example

If input string is “10111000001011” the answer will 12 as follows −

• First segment is of length 7 10111000001011
• Second segment is of length 5 10111000001011
• Total length is length of (segment 1 + segment 2) = (7 + 5) = 12

## Algorithm

• If start == n then return 0.
• Run a loop from start till n, computing for each subarray till n.
• If character is 1 then increment the count of 1 else increment the count of 0.
• If count of 1 is greater than 0, recursively call the function for index (k+1) i.e. next index and add the remaining length i.e. k-start+1.
• Else only recursively call the function for next index k+1.
• Return dp[start].

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getSegmentWithMaxLength(int start, string str, int n, int dp[]) {
if (start == n) {
return 0;
}
if (dp[start] != -1) {
return dp[start];
}
dp[start] = 0;
int one = 0;
int zero = 0;
int k;
for (k = start; k < n; ++k) {
if (str[k] == '1') {
++one;
} else {
++zero;
} if (one > zero) {
dp[start] = max(dp[start], getSegmentWithMaxLength(k + 1, str, n, dp) + k - start + 1);
} else {
dp[start] = max(dp[start], getSegmentWithMaxLength(k + 1, str, n, dp));
}
}
return dp[start];
}
int main() {
string str = "10111000001011";
int n = str.size();
int dp[n + 1];
memset(dp, -1, sizeof(dp));
cout << "Maximum length of segment = " << getSegmentWithMaxLength(0, str, n, dp) << endl;
return 0;
}

## Output

When you compile and execute above program. It generates following output −

Maximum length of segment = 12

Updated on: 10-Jan-2020

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