Given a binary string s, the task is to find the length of the longest subsequence that can be divided into three substrings such that the first and third substrings are either empty or filled with 1 and the substring at the middle is either empty or filled with 0.
Examples:
Input: s = “1001”
Output: 4
Explanation:
The entire string can be divided into the desired three parts: “1”, “00”, “1”.Input: s = “010”
Output: 2
Explanation:
The subsequences “00”, “01” and “10” can be split into three desired parts {“”, “00”, “”}, {“”, “0”, “1”} and {“1”, “0”, “”}
Approach:
To solve the problem, we need to follow the steps given below:
- Firstly, pre-compute and store in prefix arrays, the occurrences of ‘1’ and ‘0’ respectively.
- Initialize two integers i and j, where i will be the point of partition between the first and second string and j will be the point of partition between the second and third strings.
- Iterate over all possible values of i & j (0 <= i < j <=n) and find the maximum possible length of the subsequence possible which satisfies the given condition.
Below is the implementation of the above approach:
C++
// C++ Program to find the// longest subsequence possible// that starts and ends with 1// and filled with 0 in the middle#include <bits/stdc++.h>using namespace std;int longestSubseq(string s, int length){ // Prefix array to store the // occurrences of '1' and '0' int ones[length + 1], zeroes[length + 1]; // Initialise prefix arrays with 0 memset(ones, 0, sizeof(ones)); memset(zeroes, 0, sizeof(zeroes)); // Iterate over the length of the string for (int i = 0; i < length; i++) { // If current character is '1' if (s[i] == '1') { ones[i + 1] = ones[i] + 1; zeroes[i + 1] = zeroes[i]; } // If current character is '0' else { zeroes[i + 1] = zeroes[i] + 1; ones[i + 1] = ones[i]; } } int answer = INT_MIN; int x = 0; for (int i = 0; i <= length; i++) { for (int j = i; j <= length; j++) { // Add '1' available for // the first string x += ones[i]; // Add '0' available for // the second string x += (zeroes[j] - zeroes[i]); // Add '1' available for // the third string x += (ones[length] - ones[j]); // Update answer answer = max(answer, x); x = 0; } } // Print the final result cout << answer << endl;}// Driver Codeint main(){ string s = "10010010111100101"; int length = s.length(); longestSubseq(s, length); return 0;} |
Java
// Java program to find the// longest subsequence possible// that starts and ends with 1// and filled with 0 in the middleimport java.io.*;class GFG{static void longestSubseq(String s, int length){ // Prefix array to store the // occurrences of '1' and '0' int[] ones = new int[length + 1]; int[] zeroes = new int[length + 1]; // Iterate over the length of // the string for(int i = 0; i < length; i++) { // If current character is '1' if (s.charAt(i) == '1') { ones[i + 1] = ones[i] + 1; zeroes[i + 1] = zeroes[i]; } // If current character is '0' else { zeroes[i + 1] = zeroes[i] + 1; ones[i + 1] = ones[i]; } } int answer = Integer.MIN_VALUE; int x = 0; for(int i = 0; i <= length; i++) { for(int j = i; j <= length; j++) { // Add '1' available for // the first string x += ones[i]; // Add '0' available for // the second string x += (zeroes[j] - zeroes[i]); // Add '1' available for // the third string x += (ones[length] - ones[j]); // Update answer answer = Math.max(answer, x); x = 0; } } // Print the final result System.out.println(answer);}// Driver codepublic static void main(String[] args){ String s = "10010010111100101"; int length = s.length(); longestSubseq(s, length);}}// This code is contributed by offbeat |
Python3
# Python3 program to find the # longest subsequence possible # that starts and ends with 1 # and filled with 0 in the middle import sysdef longestSubseq(s, length): # Prefix array to store the # occurrences of '1' and '0' # Initialise prefix arrays with 0 ones = [0 for i in range(length + 1)] zeroes = [0 for i in range(length + 1)] # Iterate over the length of the string for i in range(length): # If current character is '1' if(s[i] == '1'): ones[i + 1] = ones[i] + 1 zeroes[i + 1] = zeroes[i] # If current character is '0' else: zeroes[i + 1] = zeroes[i] + 1 ones[i + 1] = ones[i] answer = -sys.maxsize - 1 x = 0 for i in range(length + 1): for j in range(i, length + 1): # Add '1' available for # the first string x += ones[i] # Add '0' available for # the second string x += (zeroes[j] - zeroes[i]) # Add '1' available for # the third string x += (ones[length] - ones[j]) # Update answer answer = max(answer, x) x = 0 # Print the final result print(answer)# Driver CodeS = "10010010111100101"length = len(S)longestSubseq(S, length)# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to find the// longest subsequence possible// that starts and ends with 1// and filled with 0 in the middleusing System;class GFG{static void longestSubseq(String s, int length){ // Prefix array to store the // occurrences of '1' and '0' int[] ones = new int[length + 1]; int[] zeroes = new int[length + 1]; // Iterate over the length of // the string for(int i = 0; i < length; i++) { // If current character is '1' if (s[i] == '1') { ones[i + 1] = ones[i] + 1; zeroes[i + 1] = zeroes[i]; } // If current character is '0' else { zeroes[i + 1] = zeroes[i] + 1; ones[i + 1] = ones[i]; } } int answer = int.MinValue; int x = 0; for(int i = 0; i <= length; i++) { for(int j = i; j <= length; j++) { // Add '1' available for // the first string x += ones[i]; // Add '0' available for // the second string x += (zeroes[j] - zeroes[i]); // Add '1' available for // the third string x += (ones[length] - ones[j]); // Update answer answer = Math.Max(answer, x); x = 0; } } // Print the readonly result Console.WriteLine(answer);}// Driver codepublic static void Main(String[] args){ String s = "10010010111100101"; int length = s.Length; longestSubseq(s, length);}}// This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript program to find the // longest subsequence possible // that starts and ends with 1 // and filled with 0 in the middle function longestSubseq(s, len) { // Prefix array to store the // occurrences of '1' and '0' var ones = new Array(len + 1).fill(0); var zeroes = new Array(len + 1).fill(0); // Iterate over the length of // the string for (var i = 0; i < len; i++) { // If current character is '1' if (s[i] === "1") { ones[i + 1] = ones[i] + 1; zeroes[i + 1] = zeroes[i]; } // If current character is '0' else { zeroes[i + 1] = zeroes[i] + 1; ones[i + 1] = ones[i]; } } var answer = -2147483648; var x = 0; for (var i = 0; i <= len; i++) { for (var j = i; j <= len; j++) { // Add '1' available for // the first string x += ones[i]; // Add '0' available for // the second string x += zeroes[j] - zeroes[i]; // Add '1' available for // the third string x += ones[len] - ones[j]; // Update answer answer = Math.max(answer, x); x = 0; } } // Print the readonly result document.write(answer); } // Driver code var s = "10010010111100101"; var len = s.length; longestSubseq(s, len);</script> |
Output:
12
Time Complexity: O(N2)
Auxiliary Space: O(N)
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