Given a binary string S of size N, the task is to check if the binary string S can be sorted in decreasing order by removing any number of the non-adjacent characters. If it is possible to sort the string in decreasing order, then print “Yes”. Otherwise, print “No”.
Examples:
Input: S = “10101011011”
Output: Yes
Explanation:
After removing the non-adjacent characters at indices 1, 3, 5, and 8 modifies the string to “1111111”, which is sorted in decreasing order. Therefore, print “Yes”.Input: S = “0011”
Output: No
Approach: The given problem can be solved based on the observations that if there exist two adjacent characters having 1s and then there exists adjacent characters having 0s then it is impossible to sort the string by removing the non-adjacent characters. Follow the steps below to solve the problem:
- Initialize a boolean variable, say flag as true that stores the status whether the given string can be sorted or not.
- Traverse the given string S from the end and if there exists any pairs of adjacent characters having values 1s then stored the second index of 1 in a variable say idx and break out of the loop.
- Traverse the given string S again from the back over the range [idx, 0] and if there exists any pairs of adjacent characters having values 0s then update the value of flag as false and break out of the loop.
- After completing the above steps, if the value of flag is true, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to sort the given string in// decreasing order by removing the non// adjacent charactersstring canSortString(string S, int N){ // Keeps the track whether the // string can be sorted or not int flag = 1; int i, j; // Traverse the given string S for (i = N - 2; i >= 0; i--) { // Check if S[i] and // S[i + 1] are both '1' if (S[i] == '1' && S[i + 1] == '1') { break; } } // Traverse the string S from // the indices i to 0 for (int j = i; j >= 0; j--) { // If S[j] and S[j + 1] is // equal to 0 if (S[j] == '0' && S[j + 1] == '0') { // Mark flag false flag = 0; break; } } // If flag is 0, then it is not // possible to sort the string if (flag == 0) { return "No"; } // Otherwise else { return "Yes"; }}// Driver Codeint main(){ string S = "10101011011"; int N = S.length(); cout << canSortString(S, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to sort the given string in// decreasing order by removing the non// adjacent charactersstatic String canSortString(String S, int N){ // Keeps the track whether the // string can be sorted or not int flag = 1; int i, j; // Traverse the given string S for (i = N - 2; i >= 0; i--) { // Check if S[i] and // S[i + 1] are both '1' if (S.charAt(i) == '1' && S.charAt(i + 1) == '1') { break; } } // Traverse the string S from // the indices i to 0 for ( j = i; j >= 0; j--) { // If S[j] and S[j + 1] is // equal to 0 if (S.charAt(j) == '0' && S.charAt(j + 1) == '0') { // Mark flag false flag = 0; break; } } // If flag is 0, then it is not // possible to sort the string if (flag == 0) { return "No"; } // Otherwise else { return "Yes"; }} public static void main (String[] args) { String S = "10101011011"; int N = S.length(); System.out.println(canSortString(S, N)); }}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Function to sort the given string in# decreasing order by removing the non# adjacent charactersdef canSortString(S, N): # Keeps the track whether the # string can be sorted or not flag = 1 # Traverse the given string S i = N - 2 while(i >= 0): # Check if S[i] and # S[i + 1] are both '1' if (S[i] == '1' and S[i + 1] == '1'): break i -= 1 # Traverse the string S from # the indices i to 0 j = i while(j >= 0): # If S[j] and S[j + 1] is # equal to 0 if (S[j] == '0' and S[j + 1] == '0'): # Mark flag false flag = 0 break j -= 1 # If flag is 0, then it is not # possible to sort the string if (flag == 0): return "No" # Otherwise else: return "Yes"# Driver Codeif __name__ == '__main__': S = "10101011011" N = len(S) print(canSortString(S, N))# This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;class GFG{// Function to sort the given string in// decreasing order by removing the non// adjacent charactersstatic string canSortString(string S, int N){ // Keeps the track whether the // string can be sorted or not int flag = 1; int i, j; // Traverse the given string S for(i = N - 2; i >= 0; i--) { // Check if S[i] and // S[i + 1] are both '1' if (S[i] == '1' && S[i + 1] == '1') { break; } } // Traverse the string S from // the indices i to 0 for(j = i; j >= 0; j--) { // If S[j] and S[j + 1] is // equal to 0 if (S[j] == '0' && S[j + 1] == '0') { // Mark flag false flag = 0; break; } } // If flag is 0, then it is not // possible to sort the string if (flag == 0) { return "No"; } // Otherwise else { return "Yes"; }}// Driver codepublic static void Main(string[] args){ string S = "10101011011"; int N = S.Length; Console.WriteLine(canSortString(S, N));}}// This code is contributed by ukasp |
Javascript
<script>// Javascript program for the above approach// Function to sort the given string in// decreasing order by removing the non// adjacent charactersfunction canSortString(S, N){ // Keeps the track whether the // string can be sorted or not let flag = 1; let i, j; // Traverse the given string S for (let i = N - 2; i >= 0; i--) { // Check if S[i] and // S[i + 1] are both '1' if (S[i] == '1' && S[i + 1] == '1') { break; } } // Traverse the string S from // the indices i to 0 for (let j = i; j >= 0; j--) { // If S[j] and S[j + 1] is // equal to 0 if (S[j] == '0' && S[j + 1] == '0') { // Mark flag false flag = 0; break; } } // If flag is 0, then it is not // possible to sort the string if (flag == 0) { return "No"; } // Otherwise else { return "Yes"; }}// Driver Code let S = "10101011011"; let N = S.length; document.write(canSortString(S, N));// This code is contributed by gfgking</script> |
Output:
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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