Given a binary string of 0s and 1s. The task is to find the length of substring which is having maximum difference of number of 0s and number of 1s (number of 0s – number of 1s). In case of all 1s print -1.
Examples:
Input : S = "11000010001" Output : 6 From index 2 to index 9, there are 7 0s and 1 1s, so number of 0s - number of 1s is 6. Input : S = "1111" Output : -1
The idea is to use Dynamic Programming to solve the problem.
Before that we will convert given binary string into integer array of value 1s and -1s, say arr[]. That can be easily done by traversing the given binary string and if ith index contain ‘0’ make -1 in corresponding position in array. Similarly, if ith index contain ‘1’, make 1 in the array.
Now, at each index i we need to make decision whether to take it or skip it. So, declare a 2D array of size n x 2, where n is the length of the given binary string, say dp[n][2].
dp[i][0] define the maximum value upto index i, when we skip the i-th index element. dp[i][1] define the maximum value upto index i after taking the i-th index element. Therefore, we can derive dp[i][] as: dp[i][0] = max(dp[i+1][0], dp[i+1][1] + arr[i]) dp[i][1] = max(dp[i+1][1] + arr[i], 0)
For all ones we check this case explicitly.
C++
// CPP Program to find the length of // substring with maximum difference of // zeroes and ones in binary string. #include <bits/stdc++.h> #define MAX 100 using namespace std; // Return true if there all 1s bool allones(string s, int n) { // Checking each index is 0 or not. int co = 0; for ( int i = 0; i < s.size(); i++) co += (s[i] == '1' ); return (co == n); } // Find the length of substring with maximum // difference of zeroes and ones in binary // string int findlength( int arr[], string s, int n, int ind, int st, int dp[][3]) { // If string is over. if (ind >= n) return 0; // If the state is already calculated. if (dp[ind][st] != -1) return dp[ind][st]; if (st == 0) return dp[ind][st] = max(arr[ind] + findlength(arr, s, n, ind + 1, 1, dp), findlength(arr, s, n, ind + 1, 0, dp)); else return dp[ind][st] = max(arr[ind] + findlength(arr, s, n, ind + 1, 1, dp), 0); } // Returns length of substring which is // having maximum difference of number // of 0s and number of 1s int maxLen(string s, int n) { // If all 1s return -1. if (allones(s, n)) return -1; // Else find the length. int arr[MAX] = { 0 }; for ( int i = 0; i < n; i++) arr[i] = (s[i] == '0' ? 1 : -1); int dp[MAX][3]; memset (dp, -1, sizeof dp); return findlength(arr, s, n, 0, 0, dp); } // Driven Program int main() { string s = "11000010001" ; int n = 11; cout << maxLen(s, n) << endl; return 0; } |
Java
// Java Program to find the length of // substring with maximum difference of // zeroes and ones in binary string. import java.util.Arrays; class GFG { static final int MAX= 100 ; // Return true if there all 1s static boolean allones(String s, int n) { // Checking each index is 0 or not. int co = 0 ; for ( int i = 0 ; i < s.length(); i++) if (s.charAt(i) == '1' ) co += 1 ; return (co == n); } // Find the length of substring with maximum // difference of zeroes and ones in binary // string static int findlength( int arr[], String s, int n, int ind, int st, int dp[][]) { // If string is over. if (ind >= n) return 0 ; // If the state is already calculated. if (dp[ind][st] != - 1 ) return dp[ind][st]; if (st == 0 ) return dp[ind][st] = Math.max(arr[ind] + findlength(arr, s, n, ind + 1 , 1 , dp), findlength(arr, s, n, ind + 1 , 0 , dp)); else return dp[ind][st] = Math.max(arr[ind] + findlength(arr, s, n, ind + 1 , 1 , dp), 0 ); } // Returns length of substring which is // having maximum difference of number // of 0s and number of 1s static int maxLen(String s, int n) { // If all 1s return -1. if (allones(s, n)) return - 1 ; // Else find the length. int arr[] = new int [MAX]; for ( int i = 0 ; i < n; i++) arr[i] = (s.charAt(i) == '0' ? 1 : - 1 ); int dp[][] = new int [MAX][ 3 ]; for ( int [] row : dp) Arrays.fill(row, - 1 ); return findlength(arr, s, n, 0 , 0 , dp); } // Driver code public static void main (String[] args) { String s = "11000010001" ; int n = 11 ; System.out.println(maxLen(s, n)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python Program to find the length of # substring with maximum difference of # zeroes and ones in binary string. MAX = 100 # Return true if there all 1s def allones(s, n): # Checking each index # is 0 or not. co = 0 for i in s: co + = 1 if i = = '1' else 0 return co = = n # Find the length of substring with # maximum difference of zeroes and # ones in binary string def findlength(arr, s, n, ind, st, dp): # If string is over if ind > = n: return 0 # If the state is already calculated. if dp[ind][st] ! = - 1 : return dp[ind][st] if not st: dp[ind][st] = max (arr[ind] + findlength(arr, s, n, ind + 1 , 1 , dp), (findlength(arr, s, n, ind + 1 , 0 , dp))) else : dp[ind][st] = max (arr[ind] + findlength(arr, s, n, ind + 1 , 1 , dp), 0 ) return dp[ind][st] # Returns length of substring which is # having maximum difference of number # of 0s and number of 1s def maxLen(s, n): # If all 1s return -1. if allones(s, n): return - 1 # Else find the length. arr = [ 0 ] * MAX for i in range (n): arr[i] = 1 if s[i] = = '0' else - 1 dp = [[ - 1 ] * 3 for _ in range ( MAX )] return findlength(arr, s, n, 0 , 0 , dp) # Driven Program s = "11000010001" n = 11 print (maxLen(s, n)) # This code is contributed by Ansu Kumari. |
C#
// C# Program to find the length of // substring with maximum difference of // zeroes and ones in binary string. using System; class GFG { static int MAX = 100; // Return true if there all 1s public static bool allones( string s, int n) { // Checking each index is 0 or not. int co = 0; for ( int i = 0; i < s.Length; i++) co += (s[i] == '1' ? 1 : 0); return (co == n); } // Find the length of substring with maximum // difference of zeroes and ones in binary // string public static int findlength( int [] arr, string s, int n, int ind, int st, int [,] dp) { // If string is over. if (ind >= n) return 0; // If the state is already calculated. if (dp[ind,st] != -1) return dp[ind,st]; if (st == 0) return dp[ind,st] = Math.Max(arr[ind] + findlength(arr, s, n, ind + 1, 1, dp), findlength(arr, s, n, ind + 1, 0, dp)); else return dp[ind,st] = Math.Max(arr[ind] + findlength(arr, s, n, ind + 1, 1, dp), 0); } // Returns length of substring which is // having maximum difference of number // of 0s and number of 1s public static int maxLen( string s, int n) { // If all 1s return -1. if (allones(s, n)) return -1; // Else find the length. int [] arr = new int [MAX]; for ( int i = 0; i < n; i++) arr[i] = (s[i] == '0' ? 1 : -1); int [,] dp = new int [MAX,3]; for ( int i = 0; i < MAX; i++) for ( int j = 0; j < 3; j++) dp[i,j] = -1; return findlength(arr, s, n, 0, 0, dp); } // Driven Program static void Main() { string s = "11000010001" ; int n = 11; Console.Write(maxLen(s, n)); } // This code is contributed by DrRoot_ } |
Output :
6
Maximum difference of zeros and ones in binary string | Set 2 (O(n) time)
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