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Convert given Binary string S to all 1s by changing all 0s to 1s in range [i+1, i+K] if S[i] is 1

  • Last Updated : 07 Jul, 2021

Given a binary string S of size N and a number K, the task is to find if all the ‘0’s can be changed into ‘1′s in any number of operations. In one operation, if S[i] is initially ‘1’, then all ‘0‘s in the range [i+1, i+K] can be changed to ‘1’s, for 0≤i<N-K.

Examples:

Input: S=”100100″, N=6, K=2
Output: YES
Explanation: S[0] can be used to change S[1] and S[2] into 1s
S[3] can be used to change S[4] and S[5] into 1s

Input: S=”110000″, N=6, K=2
Output: NO

 

Naive Approach: The simplest approach is to traverse the string in a reverse manner and if ‘0’ is encountered, check if the position of the nearest ‘1’ on the left is more than K indices away. If true, then print “NO”.



Time Complexity: O(N*K)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use a stack. Follow the steps below to solve the problem:

  • Initialize two variables flag and count as 1 and 0 respectively to store the result and to count the number of ‘0′s that have been changed by the last occurrence of ‘1′.
  • Initialize an empty stack st.
  • Traverse the string S, and do the following:
    • If stack is empty:
      • If the current element is ‘0’, change flag to 0 and break, as this ‘0′ cannot be changed to ‘1’.
      • Otherwise, update count to 0 and push 1 to st.
    • Otherwise:
      • If the current element is ‘0’, do the following:
        • Increment count by 1.
        • If count becomes equal to K, pop the stack st and update count to 0
      • Else, update count to 0.
  • If the value of flag is 1, print “YES”, otherwise print “NO” as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether all 0s
// in the string can be changed into 1s
void changeCharacters(string S, int N, int K)
{
    int flag = 1;
 
    // Store the count of 0s converted
    // for the last occurrence of 1
    int count = 0;
 
    // Declere a stack
    stack<char> st;
 
    // Traverse the string, S
    for (int i = 0; i < N; i++) {
 
        // If stack is empty
        if (st.empty()) {
 
            // There is no 1 that can
            // change this 0 to 1
            if (S[i] == '0') {
                flag = 0;
                break;
            }
 
            // Push 1 into the stack
            count = 0;
            st.push(S[i]);
        }
        else {
            if (S[i] == '0') {
                count++;
 
                // The last 1 has reached
                // its limit
                if (count == K) {
                    st.pop();
                    count = 0;
                }
            }
 
            // New 1 has been found which
            // can now change at most K 0s
            else {
                count = 0;
            }
        }
    }
 
    // If flag is 1, print "YES"
    // else print "NO"
    if (flag)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}
 
// Driver code
int main()
{
    // Given Input
    string S = "100100";
    int N = S.length();
    int K = 2;
 
    // Function call
    changeCharacters(S, N, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check whether all 0s
// in the string can be changed into 1s
static void changeCharacters(String S, int N, int K)
{
    int flag = 1;
 
    // Store the count of 0s converted
    // for the last occurrence of 1
    int count = 0;
 
    // Declere a stack
    Stack<Character> st = new Stack<>(); 
 
    // Traverse the string, S
    for(int i = 0; i < N; i++)
    {
         
        // If stack is empty
        if (st.empty())
        {
             
            // There is no 1 that can
            // change this 0 to 1
            if (S.charAt(i) == '0')
            {
                flag = 0;
                break;
            }
 
            // Push 1 into the stack
            count = 0;
            st.push(S.charAt(i));
        }
        else
        {
            if (S.charAt(i) == '0')
            {
                count++;
                 
                // The last 1 has reached
                // its limit
                if (count == K)
                {
                    st.pop();
                    count = 0;
                }
            }
 
            // New 1 has been found which
            // can now change at most K 0s
            else
            {
                count = 0;
            }
        }
    }
 
    // If flag is 1, print "YES"
    // else print "NO"
    if (flag == 1)
        System.out.print("YES");
    else
        System.out.print("NO");
}
 
// Driver code
public static void main(String args[])
{
     
    // Given Input
    String S = "100100";
    int N = S.length();
    int K = 2;
 
    // Function call
    changeCharacters(S, N, K);
}
}
 
// This code is contributed by ipg2016107

Python3




# Python3 program for the above approach
 
# Function to check whether all 0s
# in the string can be changed into 1s
def changeCharacters(S, N, K):
    flag = 1
 
    # Store the count of 0s converted
    # for the last occurrence of 1
    count = 0
 
    # Declere a stack
    st = []
 
    # Traverse the string, S
    for i in range(N):
 
        # If stack is empty
        if len(st) == 0:
 
            # There is no 1 that can
            # change this 0 to 1
            if (S[i] == '0'):
                flag = 0
                break
 
            # Push 1 into the stack
            count = 0
            st.append(S[i])
        else:
            if (S[i] == '0'):
                count+=1
 
                # The last 1 has reached
                # its limit
                if (count == K):
                    del st[-1]
                    count = 0
 
            # New 1 has been found which
            # can now change at most K 0s
            else:
                count = 0
 
    # If flag is 1, pr"YES"
    # else pr"NO"
    if (flag):
        print("YES")
    else:
        print("NO")
 
# Driver code
if __name__ == '__main__':
    # Given Input
    S = "100100"
    N = len(S)
    K = 2
 
    # Function call
    changeCharacters(S, N, K)
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to check whether all 0s
// in the string can be changed into 1s
static void changeCharacters(string S, int N, int K)
{
    int flag = 1;
 
    // Store the count of 0s converted
    // for the last occurrence of 1
    int count = 0;
 
    // Declere a stack
    Stack<char> st = new Stack<char>();
 
    // Traverse the string, S
    for(int i = 0; i < N; i++) {
 
        // If stack is empty
        if (st.Count==0) {
 
            // There is no 1 that can
            // change this 0 to 1
            if (S[i] == '0') {
                flag = 0;
                break;
            }
 
            // Push 1 into the stack
            count = 0;
            st.Push(S[i]);
        }
        else {
            if (S[i] == '0') {
                count++;
 
                // The last 1 has reached
                // its limit
                if (count == K) {
                    st.Pop();
                    count = 0;
                }
            }
 
            // New 1 has been found which
            // can now change at most K 0s
            else {
                count = 0;
            }
        }
    }
 
    // If flag is 1, print "YES"
    // else print "NO"
    if (flag == 1)
       Console.Write("YES");
    else
       Console.Write("NO");
}
 
// Driver code
public static void Main()
{
    // Given Input
    string S = "100100";
    int N = S.Length;
    int K = 2;
 
    // Function call
    changeCharacters(S, N, K);
 
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript




<script>
 
// JavaScript program for the above approach
 
 
// Function to check whether all 0s
// in the string can be changed into 1s
function changeCharacters(S, N, K) {
    let flag = 1;
 
    // Store the count of 0s converted
    // for the last occurrence of 1
    let count = 0;
 
    // Declere a stack
    let st = new Array();
 
    // Traverse the string, S
    for (let i = 0; i < N; i++) {
 
        // If stack is empty
        if (st.length == 0) {
 
            // There is no 1 that can
            // change this 0 to 1
            if (S[i] == '0') {
                flag = 0;
                break;
            }
 
            // Push 1 into the stack
            count = 0;
            st.push(S[i]);
        }
        else {
            if (S[i] == '0') {
                count++;
 
                // The last 1 has reached
                // its limit
                if (count == K) {
                    st.pop();
                    count = 0;
                }
            }
 
            // New 1 has been found which
            // can now change at most K 0s
            else {
                count = 0;
            }
        }
    }
 
    // If flag is 1, print "YES"
    // else print "NO"
    if (flag == 1)
        document.write("YES");
    else
        document.write("NO");
}
 
// Driver code
 
// Given Input
let S = "100100";
let N = S.Length;
let K = 2;
 
// Function call
changeCharacters(S, N, K);
 
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output
YES

Time Complexity: O(N)
Auxiliary Space: O(1) since at most one character is present in the stack at any moment

 




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