Given a binary string S of length N, the task is to find lexicographically the smallest string formed after modifying the string by selecting any substring “10” and removing any one of the characters from that substring, any number of times.
Examples:
Input: S = “0101”
Output: 001
Explanation:
Removing the S[1](=1) from the substring, “10” over the range [1, 2] modifies the string S to “001”, which is the smallest.Input: S =”11001101″
Output: 0001
Explanation:
One possible way to obtain Lexicographically the smallest string is:
- Removing the S[1](=1) from the substring, “10” over the range [1, 2] modifies the string S as S = “1001101”.
- Removing the S[0](=1) from the substring, “10” over the range [0, 1] modifies the string S as S = “001101”.
- Removing the S[3](=1) from the substring, “10” over the range [3, 4] modifies the string S as S = “00101”.
- Removing the S[2](=1) from the substring, “10” over the range [2, 3] modifies the string S as S = “0001”.
- Now any character can not be removed.
Therefore, the lexicographically smallest obtained string is “0001”.
Approach: The given problem can be solved based on the following observations:
- It can be observed that the character ‘1‘ after the last zero can not be removed as one will not be able to find any “10” substring.
- Lexicographically, the smallest string will contain as much as zero as it can before the first one.
- It can be observed that every ‘1‘ can be deleted if there is at least one ‘0‘ after it.
- Therefore, the idea is to remove all the ones before the last occurring ‘0‘ from the string.
Follow the steps below to solve the problem:
- Initialize two variables, say ans and LastZe, to store the resulting string and the index of last occurring ‘0‘.
- Iterate over the characters of string S using the variable i and then if S[i] is ‘0‘ then assign i to LastZe.
- Iterate over the characters of string S using the variable i and perform the following operations:
- If S[i] = ‘0’ and i ? LastZe, then append S[i] to ans.
- Otherwise, if i > LastZe, then append S[i] to ans.
- Finally, after completing the above steps, print the result as ans.
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 smallest lexicogra-// phically smallest stringstring lexicographicallySmallestString(string S, int N){ // Stores the index of last // occurring 0 int LastZe = -1; // Stores the lexicographically // smallest string string ans; // Traverse the string S for (int i = N - 1; i >= 0; i--) { // If str[i] is 0 if (S[i] == '0') { // Assign i to lastZe LastZe = i; break; } } // Traverse the string str for (int i = 0; i < N; i++) { // If i is less than or equal // to lastZe and str[i] is 0 if (i <= LastZe && S[i] == '0') ans += S[i]; // If i is greater than lastZe else if (i > LastZe) ans += S[i]; } // Return ans return ans;}// Driver Codeint main(){ // Input string S = "11001101"; int N = S.size(); // Function Call cout << lexicographicallySmallestString(S, N); return 0;} |
Java
// Java program for the above approachimport java.lang.*;import java.util.*;class GFG{// Function to find smallest lexicogra-// phically smallest stringstatic String lexicographicallySmallestString(String S, int N){ // Stores the index of last // occurring 0 int LastZe = -1; // Stores the lexicographically // smallest string String ans = ""; // Traverse the string S for(int i = N - 1; i >= 0; i--) { // If str[i] is 0 if (S.charAt(i) == '0') { // Assign i to lastZe LastZe = i; break; } } // Traverse the string str for(int i = 0; i < N; i++) { // If i is less than or equal // to lastZe and str[i] is 0 if (i <= LastZe && S.charAt(i) == '0') ans += S.charAt(i); // If i is greater than lastZe else if (i > LastZe) ans += S.charAt(i); } // Return ans return ans;}// Driver codepublic static void main(String[] args){ // Input String S = "11001101"; int N = S.length(); // Function Call System.out.println( lexicographicallySmallestString(S, N));}}// This code is contributed by avijitmondal1998 |
Python3
# Python program for the above approach# Function to find smallest lexicogra-# phically smallest stringdef lexicographicallySmallestString(S, N): # Stores the index of last # occurring 0 LastZe = -1 # Stores the lexicographically # smallest string ans = "" # Traverse the S for i in range(N - 1, -1, -1): # If str[i] is 0 if (S[i] == '0'): # Assign i to lastZe LastZe = i break # Traverse the str for i in range(N): # If i is less than or equal # to lastZe and str[i] is 0 if (i <= LastZe and S[i] == '0'): ans += S[i] # If i is greater than lastZe elif (i > LastZe): ans += S[i] # Return ans return ans# Driver Codeif __name__ == '__main__': # Input S = "11001101" N = len(S) # Function Call print (lexicographicallySmallestString(S, N))# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;class GFG { // Function to find smallest lexicogra- // phically smallest string static string lexicographicallySmallestString(string S, int N) { // Stores the index of last // occurring 0 int LastZe = -1; // Stores the lexicographically // smallest string string ans = ""; // Traverse the string S for (int i = N - 1; i >= 0; i--) { // If str[i] is 0 if (S[i] == '0') { // Assign i to lastZe LastZe = i; break; } } // Traverse the string str for (int i = 0; i < N; i++) { // If i is less than or equal // to lastZe and str[i] is 0 if (i <= LastZe && S[i] == '0') ans += S[i]; // If i is greater than lastZe else if (i > LastZe) ans += S[i]; } // Return ans return ans; } // Driver Code public static void Main() { // Input string S = "11001101"; int N = S.Length; // Function Call Console.Write( lexicographicallySmallestString(S, N)); }}// This code is contributed by ukasp. |
Javascript
<script>// JavaScript program for the above approach// Function to find smallest lexicogra-// phically smallest stringfunction lexicographicallySmallestString(S, N){ // Stores the index of last // occurring 0 var LastZe = -1; // Stores the lexicographically // smallest string var ans = ""; // Traverse the string S for(var i = N - 1; i >= 0; i--) { // If str[i] is 0 if (S.charAt(i) == '0') { // Assign i to lastZe LastZe = i; break; } } // Traverse the string str for(var i = 0; i < N; i++) { // If i is less than or equal // to lastZe and str[i] is 0 if (i <= LastZe && S.charAt(i) == '0') ans += S.charAt(i); // If i is greater than lastZe else if (i > LastZe) ans += S.charAt(i); } // Return ans return ans;}// Driver code// Inputvar S = "11001101";var N = S.length;// Function Calldocument.write(lexicographicallySmallestString(S, N));// This code is contributed by shivanisinghss2110</script> |
Output
0001
Time Complexity: O(N)
Auxiliary Space: O(1)
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