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:
- 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 0svoid 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 Codeint main(){ int K1 = 2, K2 = 4; string S = "11000100"; generateString(K1, K2, S); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find lexicographically// smallest String having number of 1s// greater than number of 0sstatic 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 Codepublic 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 0sdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG {// Function to find lexicographically// smallest String having number of 1s// greater than number of 0sstatic 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
