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Lexicographically smallest binary string formed by flipping bits at indices not divisible K1 or K2 such that count of 1s is always greater than 0s from left

  • Last Updated : 05 Oct, 2021
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Given a binary string S(1-based indexing) of size N, and two positive integers K1 and K2, the task is to find the lexicographically smallest string by flipping the characters at indices that are not divisible by either K1 or K2 such that the count of 1s till every possible index is always greater than the count of 0s. If it is not possible to form such string, the print “-1”.

Examples:

Input: K1 = 4, K2 = 6, S = “0000”
Output: 1110
Explanation:
Since the index 4 is divisible by K1(= 4). So without flipping that index the string modifies to  “1110”, which is lexicographically smallest among all possible combinations of flips.

Input: K1 = 2, K2 = 4, S = “11000100”
Output: 11100110

Approach: The problem can be solved by modifying the string S from left to right for every unlocked position, if it is possible to make 0 then convert it to 0 else convert it to 1. Follow the steps below to solve the problem:



  • Initialize two variables say c1 and c0 to store the count of 1s and 0s respectively.
  • Initialize a vector say pos[] that stores the positions of all the 0s that are not divisible by K1 or K2.
  • Traverse the given string S and perform the following steps:
    • If the character is 0 then increment the value of c0. Otherwise, increment the value of c1.
    • If the current index is not divisible by K1 or K2, then insert this index in the vector pos[].
    • If at any index i, the count of 0s becomes greater than or equal to 1s then:
      • If the vector is empty then, the string can’t be modified to required combination. Hence, print -1.
      • Otherwise, pop the last position present in vector and update  the value of c0 as c0 – 1 and c1 as c1 + 1 and S[pos] = ‘1’.
  • After completing the above steps, print the string S the modified string.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find lexicographically
// smallest string having number of 1s
// greater than number of 0s
void generateString(int k1, int k2, string s)
{
    // C1s And C0s stores the count of
    // 1s and 0s at every position
    int C1s = 0, C0s = 0;
    int flag = 0;
    vector<int> pos;
 
    // Traverse the string S
    for (int i = 0; i < s.length(); i++) {
 
        if (s[i] == '0') {
            C0s++;
 
            // If the position is not
            // divisible by k1 and k2
            if ((i + 1) % k1 != 0
                && (i + 1) % k2 != 0) {
                pos.push_back(i);
            }
        }
 
        else {
            C1s++;
        }
 
        if (C0s >= C1s) {
 
            // If C0s >= C1s and pos[] is
            // empty then the string can't
            // be formed
            if (pos.size() == 0) {
                cout << -1;
                flag = 1;
                break;
            }
 
            // If pos[] is not empty then
            // flip the bit of last position
            // present in pos[]
            else {
                int k = pos.back();
                s[k] = '1';
                C0s--;
                C1s++;
                pos.pop_back();
            }
        }
    }
 
    // Print the result
    if (flag == 0) {
        cout << s;
    }
}
 
// Driver Code
int main()
{
    int K1 = 2, K2 = 4;
    string S = "11000100";
    generateString(K1, K2, S);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG
{
 
// Function to find lexicographically
// smallest String having number of 1s
// greater than number of 0s
static void generateString(int k1, int k2, char[] s)
{
   
    // C1s And C0s stores the count of
    // 1s and 0s at every position
    int C1s = 0, C0s = 0;
    int flag = 0;
    Vector<Integer> pos = new Vector<Integer>();
 
    // Traverse the String S
    for (int i = 0; i < s.length; i++) {
 
        if (s[i] == '0') {
            C0s++;
 
            // If the position is not
            // divisible by k1 and k2
            if ((i + 1) % k1 != 0
                && (i + 1) % k2 != 0) {
                pos.add(i);
            }
        }
 
        else {
            C1s++;
        }
 
        if (C0s >= C1s) {
 
            // If C0s >= C1s and pos[] is
            // empty then the String can't
            // be formed
            if (pos.size() == 0) {
                System.out.print(-1);
                flag = 1;
                break;
            }
 
            // If pos[] is not empty then
            // flip the bit of last position
            // present in pos[]
            else {
                int k = pos.get(pos.size()-1);
                s[k] = '1';
                C0s--;
                C1s++;
                pos.remove(pos.size() - 1);
            }
        }
    }
 
    // Print the result
    if (flag == 0) {
        System.out.print(s);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int K1 = 2, K2 = 4;
    String S = "11000100";
    generateString(K1, K2, S.toCharArray());
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 program for the above approach
 
# Function to find lexicographically
# smallest string having number of 1s
# greater than number of 0s
def generateString(k1, k2, s):
   
    # C1s And C0s stores the count of
    # 1s and 0s at every position
    s = list(s)
    C1s = 0
    C0s = 0
    flag = 0
    pos = []
 
    # Traverse the string S
    for i in range(len(s)):
        if (s[i] == '0'):
            C0s += 1
 
            # If the position is not
            # divisible by k1 and k2
            if ((i + 1) % k1 != 0 and (i + 1) % k2 != 0):
                pos.append(i)
 
        else:
            C1s += 1
 
        if (C0s >= C1s):
            # If C0s >= C1s and pos[] is
            # empty then the string can't
            # be formed
            if (len(pos) == 0):
                print(-1)
                flag = 1
                break
 
            # If pos[] is not empty then
            # flip the bit of last position
            # present in pos[]
            else:
                k = pos[len(pos)-1]
                s[k] = '1'
                C0s -= 1
                C1s += 1
                pos = pos[:-1]
 
    # Print the result
    s = ''.join(s)
    if (flag == 0):
        print(s)
 
# Driver Code
if __name__ == '__main__':
    K1 = 2
    K2 = 4
    S = "11000100"
    generateString(K1, K2, S)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
// Function to find lexicographically
// smallest String having number of 1s
// greater than number of 0s
static void generateString(int k1, int k2, char[] s)
{
    
    // C1s And C0s stores the count of
    // 1s and 0s at every position
    int C1s = 0, C0s = 0;
    int flag = 0;
    List<int> pos = new List<int>();
  
    // Traverse the String S
    for (int i = 0; i < s.Length; i++) {
  
        if (s[i] == '0') {
            C0s++;
  
            // If the position is not
            // divisible by k1 and k2
            if ((i + 1) % k1 != 0
                && (i + 1) % k2 != 0) {
                pos.Add(i);
            }
        }
  
        else {
            C1s++;
        }
  
        if (C0s >= C1s) {
  
            // If C0s >= C1s and pos[] is
            // empty then the String can't
            // be formed
            if (pos.Count == 0) {
                Console.WriteLine(-1);
                flag = 1;
                break;
            }
  
            // If pos[] is not empty then
            // flip the bit of last position
            // present in pos[]
            else {
                int k = pos[(pos.Count - 1)];
                s[k] = '1';
                C0s--;
                C1s++;
                pos.Remove(pos.Count - 1);
            }
        }
    }
  
    // Print the result
    if (flag == 0) {
        Console.WriteLine(s);
    }
}
 
    // Driver Code
    public static void Main()
    {
    int K1 = 2, K2 = 4;
    string S = "11000100";
    generateString(K1, K2, S.ToCharArray());
    }
}
 
// This code is contributed by avijitmondal1998.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find lexicographically
        // smallest string having number of 1s
        // greater than number of 0s
        function generateString(k1, k2, s)
        {
         
            // C1s And C0s stores the count of
            // 1s and 0s at every position
            let C1s = 0, C0s = 0;
            let flag = 0;
            let pos = [];
 
            // Traverse the string S
            for (let i = 0; i < s.length; i++) {
 
                if (s[i] == '0') {
                    C0s++;
 
                    // If the position is not
                    // divisible by k1 and k2
                    if ((i + 1) % k1 != 0
                        && (i + 1) % k2 != 0) {
                        pos.push(i);
                    }
                }
 
                else {
                    C1s++;
                }
 
                if (C0s >= C1s) {
 
                    // If C0s >= C1s and pos[] is
                    // empty then the string can't
                    // be formed
                    if (pos.length == 0) {
                        cout << -1;
                        flag = 1;
                        break;
                    }
 
                    // If pos[] is not empty then
                    // flip the bit of last position
                    // present in pos[]
                    else {
                        let k = pos[pos.length - 1];
                        var ns = s.replace(s[k], '1');
                        C0s--;
                        C1s++;
                        pos.pop();
                    }
                }
            }
 
            // Print the result
            if (flag == 0) {
                document.write(ns);
            }
        }
 
        // Driver Code
        let K1 = 2, K2 = 4;
        let S = "11000100";
        generateString(K1, K2, S);
 
// This code is contributed by Potta Lokesh
 
    </script>
Output: 
11100110

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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