Given a string, S of even length N, the task is to find the minimum number of moves needed to convert the string to a string consisting of half upper case and half lower case characters.
Examples:
Input: S = “AbcdEf”
Output: 1
ABcdEf
Explanation:
One possible way to modify the string is by converting the second character to S[1]( =’b’) to an uppercase character. Thereafter, the string modifies to “ABcdEf” which consists of 3 lower case and upper case characters.Input: S = “ABCdef”
Output: 0
ABCdef
Approach: The idea is to convert only those characters that are extra in any of the cases. Follow the steps below to solve the problem:
- Find the count of the upper case English alphabet and lower case English alphabet and store them in variables, say upper and lower respectively.
- Initialize a variable, say moves as 0, to store the minimum number of moves to modify the string.
- If upper case characters are more, then traverse the string and convert upper case characters into the lower case until both cases characters are equal in number.
- Otherwise, if the lower case characters are more, traverse the string S and convert the lower case characters into the upper case until both cases characters are equal in number.
- Finally, after completing the above steps, print the value of moves as the answer and string S as 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 calculate minimum// number of moves required to// convert the stringvoid minimumTimeToConvertString(string S, int N){ // Stores Count of upper and // lower case characters int upper = 0, lower = 0; // Traverse the string S for (int i = 0; i < N; i++) { char c = S[i]; // If current character is // uppercase if (isupper(c)) { // Increment count // of Uppercase characters upper++; } // Otherwise, else { // Increment count // of Lowercase characters lower++; } } // Stores minimum number of moves needed int moves = 0; // If there are more upper // case characters if (upper > N / 2) { int i = 0; // Iterate until upper is greater // than N/2 while (upper > N / 2 && i < N) { // Convert uppercase into // lowercase until upper=N/2 if (isupper(S[i])) { S[i] += 32; moves++; upper--; lower++; } // Increment the pointer i++; } } // If there are more lower // case characters else if (lower > N / 2) { int i = 0; // Iterate until lower is greater // than N/2 while (lower > N / 2 && i < N) { // Convert lowercase into // uppercase until lower=N/2 if (islower(S[i])) { S[i] -= 32; moves++; upper++; lower--; } // Increment the pointer i++; } } // Print moves required cout << moves << endl; // Print resultant string cout << S << endl;}// Driver Codeint main(){ // Given string string S = "AbcdEf"; int N = S.length(); // Function call minimumTimeToConvertString(S, N); return 0;} |
Java
// Java program for the above approachclass GFG{// Function to calculate minimum// number of moves required to// convert the stringstatic void minimumTimeToConvertString(String S, int N){ // Stores Count of upper and // lower case characters int upper = 0, lower = 0; // Traverse the string S for(int i = 0; i < N; i++) { char c = S.charAt(i); // If current character is // uppercase if (Character.isUpperCase(c)) { // Increment count // of Uppercase characters upper++; } // Otherwise, else { // Increment count // of Lowercase characters lower++; } } // Stores minimum number of moves needed int moves = 0; // If there are more upper // case characters if (upper > N / 2) { int i = 0; // Iterate until upper is greater // than N/2 while (upper > N / 2 && i < N) { // Convert uppercase into // lowercase until upper=N/2 if (Character.isUpperCase(S.charAt(i))) { S = S.substring(0, i) + (char)(S.charAt(i) + 32) + S.substring(i + 1); moves++; upper--; lower++; } // Increment the pointer i++; } } // If there are more lower // case characters else if (lower > N / 2) { int i = 0; // Iterate until lower is greater // than N/2 while (lower > N / 2 && i < N) { // Convert lowercase into // uppercase until lower=N/2 if (Character.isLowerCase(S.charAt(i))) { S = S.substring(0, i) + (char)(S.charAt(i) - 32) + S.substring(i + 1); moves++; upper++; lower--; } // Increment the pointer i++; } } // Print moves required System.out.println(moves); // Print resultant string System.out.println(S);}// Driver codepublic static void main(String[] args){ // Given string String S = "AbcdEf"; int N = S.length(); // Function call minimumTimeToConvertString(S, N);}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to calculate minimum# number of moves required to# convert the stringdef minimumTimeToConvertString(S, N): S = [i for i in S] # Stores Count of upper and # lower case characters upper = 0 lower = 0 # Traverse the S for i in range(N): c = S[i] # If current character is # uppercase if (c.isupper()): # Increment count # of Uppercase characters upper += 1 # Otherwise, else: # Increment count # of Lowercase characters lower += 1 # Stores minimum number of moves needed moves = 0 # If there are more upper # case characters if (upper > N // 2): i = 0 # Iterate until upper is greater # than N/2 while (upper > N // 2 and i < N): # Convert uppercase into # lowercase until upper=N/2 if (S[i].isupper()): S[i] += 32 moves += 1 upper -= 1 lower += 1 # Increment the pointer i += 1 # If there are more lower # case characters elif (lower > N // 2): i = 0 # Iterate until lower is greater # than N/2 while (lower > N // 2 and i < N): # Convert lowercase into # uppercase until lower=N/2 if (S[i].islower()): S[i] = chr(ord(S[i]) - 32) moves += 1 upper += 1 lower -= 1 # Increment the pointer i += 1 # Print moves required print(moves) # Print resultant string print("".join(S))# Driver Codeif __name__ == '__main__': # Given string S = "AbcdEf" N = len(S) # Function call minimumTimeToConvertString(S, N)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG { // Function to calculate minimum // number of moves required to // convert the string static void minimumTimeToConvertString(char[] S, int N) { // Stores Count of upper and // lower case characters int upper = 0, lower = 0; // Traverse the string S for (int i = 0; i < N; i++) { char c = S[i]; // If current character is // uppercase if (Char.IsUpper(c)) { // Increment count // of Uppercase characters upper++; } // Otherwise, else { // Increment count // of Lowercase characters lower++; } } // Stores minimum number of moves needed int moves = 0; // If there are more upper // case characters if (upper > N / 2) { int i = 0; // Iterate until upper is greater // than N/2 while (upper > N / 2 && i < N) { // Convert uppercase into // lowercase until upper=N/2 if (Char.IsUpper(S[i])) { S[i] += (char)32; moves++; upper--; lower++; } // Increment the pointer i++; } } // If there are more lower // case characters else if (lower > N / 2) { int i = 0; // Iterate until lower is greater // than N/2 while (lower > N / 2 && i < N) { // Convert lowercase into // uppercase until lower=N/2 if (Char.IsLower(S[i])) { S[i] = (char)((int)S[i] - 32); moves++; upper++; lower--; } // Increment the pointer i++; } } // Print moves required Console.WriteLine(moves); // Print resultant string Console.WriteLine(new string(S)); } static void Main() { // Given string string S = "AbcdEf"; int N = S.Length; // Function call minimumTimeToConvertString(S.ToCharArray(), N); }}// This code is contributed by mukesh07. |
Javascript
<script> // Javascript program for the above approach // Function to calculate minimum // number of moves required to // convert the string function minimumTimeToConvertString(S, N) { // Stores Count of upper and // lower case characters let upper = 0, lower = 0; // Traverse the string S for (let i = 0; i < N; i++) { let c = S[i]; // If current character is // uppercase if (c == c.toUpperCase()) { // Increment count // of Uppercase characters upper++; } // Otherwise, else { // Increment count // of Lowercase characters lower++; } } // Stores minimum number of moves needed let moves = 0; // If there are more upper // case characters if (upper > parseInt(N / 2, 10)) { let i = 0; // Iterate until upper is greater // than N/2 while (upper > parseInt(N / 2, 10) && i < N) { // Convert uppercase into // lowercase until upper=N/2 if (S[i] == S[i].toUpperCase()) { S[i] += String.fromCharCode(32); moves++; upper--; lower++; } // Increment the pointer i++; } } // If there are more lower // case characters else if (lower > parseInt(N / 2, 10)) { let i = 0; // Iterate until lower is greater // than N/2 while (lower > parseInt(N / 2, 10) && i < N) { // Convert lowercase into // uppercase until lower=N/2 if (S[i] == S[i].toLowerCase()) { S[i] = String.fromCharCode(S[i].charCodeAt() - 32); moves++; upper++; lower--; } // Increment the pointer i++; } } // Print moves required document.write(moves + "</br>"); // Print resultant string document.write(S.join("")); } // Given string let S = "AbcdEf"; let N = S.length; // Function call minimumTimeToConvertString(S.split(''), N); // This code is contributed by decode2207.</script> |
Output
1 ABcdEf
Time Complexity: O(N)
Auxiliary Space: O(1)
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