Given a string S and integer M. The task is to perform exactly M operations to get lexicographical smallest string.
- In each operation, select one character optimally from the string and update it with immediate next character ( aaa -> aab ), so that string remain lexicographical smallest.
- Multiple operations are allowed over single character of the string.
Note: Consider next of ‘z’ as ‘a’.
Examples:
Input: S = “aazzx”, M = 6
Output: aaaab
Explanation:
We try to form lexicographical smallest string for each operations.
For m = 1: update “aazzx” to “aaazx”
for m = 2: update “aaazx” to “aaaax”
for m = 3: update “aaaax” to “aaaay”
for m = 4: update “aaaay” to “aaaaz”
for m = 5: update “aaaaz” to “aaaaa”
for m = 6: update “aaaaa” to “aaaab” which is lexicographical smallest than “aaaba”, “aabaa”, “abaaa”, “baaaa”.
Final string after 6 operations: “aaaab”.
Input: S = “z”, M = 27
Output: a
Explanation:
Try to form lexicographical smallest string for each operations, since there is only single character so all 27 operations have to performed over it. Final string after 27 operation is “a”.
Approach: The idea is to implement a simple greedy approach, to iterate the string from the starting of the string and while iterating focus on the current element of the string to make it as small as possible.
- Suppose, current element is r, to make r smallest i.e. a, it require 9 operations, let call the value as distance.
- Now, check if M is greater than equal to the distance, then update current character to ‘a’ and decrease the value of M by distance. Otherwise continue with next iteration.
- As next of z is a, cycle of 26 is formed. So the last character of the string, can be updated with last character + (M % 26).
Below is the implementation of the above approach:
C++
// C++ implementation to find the// lexicographical smallest string// after performing M operations#include <bits/stdc++.h>using namespace std;// Function to find the// lexicographical smallest string// after performing M operationsvoid smallest_string(string s, int m){ // Size of the given string int n = s.size(); // Declare an array a int a[n]; // For each i, a[i] contain number // of operations to update s[i] to 'a' for (int i = 0; i < n; i++) { int distance = s[i] - 'a'; if (distance == 0) a[i] = 0; else a[i] = 26 - distance; } for (int i = 0; i < n; i++) { // Check if m >= ar[i], // then update s[i] to 'a' // decrement k by a[i] if (m >= a[i]) { s[i] = 'a'; m = m - a[i]; } } // Form a cycle of 26 m = m % 26; // update last element of // string with the value // s[i] + (k % 26) s[n - 1] = s[n - 1] + m; // Return the answer cout << s;}// Driver codeint main(){ string str = "aazzx"; int m = 6; smallest_string(str, m); return 0;} |
Java
// Java implementation to find the// lexicographical smallest String// after performing M operationsclass GFG{// Function to find the// lexicographical smallest String// after performing M operationsstatic void smallest_String(char []s, int m){ // Size of the given String int n = s.length; // Declare an array a int []a = new int[n]; // For each i, a[i] contain number // of operations to update s[i] to 'a' for (int i = 0; i < n; i++) { int distance = s[i] - 'a'; if (distance == 0) a[i] = 0; else a[i] = 26 - distance; } for (int i = 0; i < n; i++) { // Check if m >= ar[i], // then update s[i] to 'a' // decrement k by a[i] if (m >= a[i]) { s[i] = 'a'; m = m - a[i]; } } // Form a cycle of 26 m = m % 26; // update last element of // String with the value // s[i] + (k % 26) s[n - 1] = (char) (s[n - 1] + m); // Return the answer System.out.print(String.valueOf(s));}// Driver codepublic static void main(String[] args){ String str = "aazzx"; int m = 6; smallest_String(str.toCharArray(), m);}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation to find the # lexicographical smallest string # after performing M operations # Function to find the# lexicographical smallest string# after performing M operationsdef smallest_string(s, m): # Size of the given string n = len(s); l = list(s) # Declare an array a a = [0] * n; # For each i, a[i] contain number # of operations to update s[i] to 'a' for i in range(n): distance = ord(s[i]) - ord('a'); if (distance == 0): a[i] = 0; else: a[i] = 26 - distance; for i in range(n): # Check if m >= ar[i], # then update s[i] to 'a' # decrement k by a[i] if (m >= a[i]): l[i] = 'a'; m = m - a[i]; # Form a cycle of 26 m = m % 26; # update last element of # with the value # s[i] + (k % 26) # Return the answer for i in range(len(l) - 1): print(l[i], end = "") print(chr(ord(l[n - 1]) + m))# Driver codestr = "aazzx";m = 6;smallest_string(str, m);# This code is contributed by grand_master |
C#
// C# implementation to find the// lexicographical smallest String// after performing M operationsusing System;class GFG{// Function to find the// lexicographical smallest String// after performing M operationsstatic void smallest_String(char []s, int m){ // Size of the given String int n = s.Length; // Declare an array a int []a = new int[n]; // For each i, a[i] contain number // of operations to update s[i] to 'a' for(int i = 0; i < n; i++) { int distance = s[i] - 'a'; if (distance == 0) a[i] = 0; else a[i] = 26 - distance; } for(int i = 0; i < n; i++) { // Check if m >= ar[i], // then update s[i] to 'a' // decrement k by a[i] if (m >= a[i]) { s[i] = 'a'; m = m - a[i]; } } // Form a cycle of 26 m = m % 26; // Update last element of // String with the value // s[i] + (k % 26) s[n - 1] = (char)(s[n - 1] + m); // Return the answer Console.Write(String.Join("", s));}// Driver codepublic static void Main(String[] args){ String str = "aazzx"; int m = 6; smallest_String(str.ToCharArray(), m);}}// This code is contributed by Princi Singh |
Javascript
<script>// JavaScript implementation to find the// lexicographical smallest string// after performing M operations// Function to find the// lexicographical smallest string// after performing M operationsfunction smallest_string(s,m){ // Size of the given string let n = s.length; // Declare an array a let a = new Array(n).fill(0); // For each i, a[i] contain number // of operations to update s[i] to 'a' for (let i = 0; i < n; i++) { let distance = s.charCodeAt(i) - 'a'.charCodeAt(0); if (distance == 0) a[i] = 0; else a[i] = 26 - distance; } for (let i = 0; i <br n; i++) { // Check if m >= ar[i], // then update s[i] to 'a' // decrement k by a[i] if (m >= a[i]) { s = s.replace(s[i],'a'); m = m - a[i]; } } // Form a cycle of 26 m = m % 26; // update last element of // string with the value // s[i] + (k % 26) s = s.substring(0,n-1) + String.fromCharCode(s.charCodeAt(n - 1) + m); // Return the answer document.write(s,"</br>");}// Driver codelet str = "aazzx";let m = 6;smallest_string(str, m)// This code is contributed by shinjanpatra</script> |
Output:
aaaab
Time Complexity: O(N), where N is the length of given string
Auxiliary Space: O(N), where N is the length of given string
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