Given a binary array, find the length of the longest subsequence such that there is no 0 after a 1.
Examples:
Input : 1 1 0 1 Output : 3 Explanation : If we remove 0 from the array, then no zero comes right after one (satisfying the condition) and the maximum game left are 3 (i.e. 1 1 1) Input : 0 Output : 1 Explanation : Since he wants to save maximum game in the array. He doesn't remove any game.
Let’s find out how many zeros will be in this sequence and then take all ones which come after the last zero. On each step take the next zero from the beginning of the sequence and count the ones after it. Update answer with the maximum value.
You can pre-calculate the number of ones on the suffix.
E.g. 0 1 0 0 1 1 1
After calculating the suffix, the array becomes :
0 4 0 0 3 2 1
Move from start to end and each time zero is found in the array increment numberofzeros by 1. If the array[index] is not zero then res = max(res, numberofzeros + value of the array at that index).
And then after the loop: res = max(res, numberofzeros)
Implementation:
C++
// CPP program to find longest subsequence// such that there is no 0 after 1.#include <bits/stdc++.h>using namespace std;int maxSubseq(int vec[], int n) { // Store the count of number of ones // from right to left in the array int suffix = 0; for (int i = n - 1; i >= 0; i--) { if (vec[i] == 1) { suffix++; vec[i] = suffix; } } // Traverse from left to right, keep count // of 0s and for every 0, check number of // 1s after it. Update the result if needed. int res = 0; int zero = 0; for (int i = 0; i < n; i++) { if (vec[i] == 0) zero++; // Checking the maximum size if (vec[i] > 0) res = max(res, zero + vec[i]); } // Checking the maximum size return max(res, zero);}// Driver Codeint main(){ int input[] = { 0, 1, 0, 0, 1, 0 }; int n = sizeof(input) / sizeof(input[0]); cout << maxSubseq(input, n); return 0;} |
Java
// java program to find longest subsequence// such that there is no 0 after 1.import java.io.*;public class GFG { static int maxSubseq(int []vec, int n) { // Store the count of number of // ones from right to left in // the array int suffix = 0; for (int i = n - 1; i >= 0; i--) { if (vec[i] == 1) { suffix++; vec[i] = suffix; } } // Traverse from left to right, keep // count of 0s and for every 0, check // number of 1s after it. Update the // result if needed. int res = 0; int zero = 0; for (int i = 0; i < n; i++) { if (vec[i] == 0) zero++; // Checking the maximum size if (vec[i] > 0) res = Math.max(res, zero + vec[i]); } // Checking the maximum size return Math.max(res, zero); } // Driver Code static public void main (String[] args) { int []input = { 0, 1, 0, 0, 1, 0 }; int n = input.length; System.out.println(maxSubseq(input, n)); }}// This code is contributed by vt_m. |
Python3
# Python 3 program to find longest subsequence# such that there is no 0 after 1.def maxSubseq(vec, n): # Store the count of number of ones # from right to left in the array suffix = 0 i = n-1 while(i >= 0): if (vec[i] == 1): suffix += 1 vec[i] = suffix i -= 1 # Traverse from left to right, keep count # of 0s and for every 0, check number of # 1s after it. Update the result if needed. res = 0 zero = 0 for i in range(0,n,1): if (vec[i] == 0): zero += 1 # Checking the maximum size if (vec[i] > 0): res = max(res, zero + vec[i]) # Checking the maximum size return max(res, zero)# Driver code if __name__ == '__main__': input = [0, 1, 0, 0, 1, 0] n = len(input) print(maxSubseq(input, n))# This code is contributed by # Surendra_Gangwar |
C#
// C# program to find longest subsequence// such that there is no 0 after 1.using System;public class GFG { static int maxSubseq(int []vec, int n) { // Store the count of number of // ones from right to left in // the array int suffix = 0; for (int i = n - 1; i >= 0; i--) { if (vec[i] == 1) { suffix++; vec[i] = suffix; } } // Traverse from left to right, keep // count of 0s and for every 0, check // number of 1s after it. Update the // result if needed. int res = 0; int zero = 0; for (int i = 0; i < n; i++) { if (vec[i] == 0) zero++; // Checking the maximum size if (vec[i] > 0) res = Math.Max(res, zero + vec[i]); } // Checking the maximum size return Math.Max(res, zero); } // Driver Code static public void Main () { int []input = { 0, 1, 0, 0, 1, 0 }; int n = input.Length; Console.WriteLine(maxSubseq(input, n)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find longest // subsequence such that there // is no 0 after 1.function maxSubseq($vec, $n) { // Store the count of // number of ones from // right to left in // the array $suffix = 0; for ($i = $n - 1; $i >= 0; $i--) { if ($vec[$i] == 1) { $suffix++; $vec[$i] = $suffix; } } // Traverse from left to // right, keep count of // 0s and for every 0, // check number of 1s after // it. Update the result if // needed. $res = 0; $zero = 0; for ($i = 0; $i < $n; $i++) { if ($vec[$i] == 0) $zero++; // Checking the // maximum size if ($vec[$i] > 0) $res = max($res, $zero + $vec[$i]); } // Checking the // maximum size return max($res, $zero);}// Driver Code$input = array(0, 1, 0, 0, 1, 0);$n = count($input); echo maxSubseq($input, $n);// This code is contributed// by anuj_67.?> |
Javascript
<script> // Javascript program to find longest subsequence // such that there is no 0 after 1. function maxSubseq(vec, n) { // Store the count of number of // ones from right to left in // the array let suffix = 0; for (let i = n - 1; i >= 0; i--) { if (vec[i] == 1) { suffix++; vec[i] = suffix; } } // Traverse from left to right, keep // count of 0s and for every 0, check // number of 1s after it. Update the // result if needed. let res = 0; let zero = 0; for (let i = 0; i < n; i++) { if (vec[i] == 0) zero++; // Checking the maximum size if (vec[i] > 0) res = Math.max(res, zero + vec[i]); } // Checking the maximum size return Math.max(res, zero); } let input = [ 0, 1, 0, 0, 1, 0 ]; let n = input.length; document.write(maxSubseq(input, n)); </script> |
Output
4
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