# Minimum Number of Operations to move all Uppercase Characters before all Lower Case Characters

You are given a string 'str' that contains both uppercase and lowercase letters. Any lowercase character can be changed to an uppercase character and vice versa in a single action. The goal is to print the least possible instances of this process that are necessary to produce a string containing at least one lowercase character, followed by at least one uppercase character.

## Input Output Scenarios

First possible solution: the first 4 characters can be converted to uppercase characters i.e. “TUTORial” with 4 operations.

Input

str = “tutoRial”


Output

1


Second possible solution: the third character can be converted to a lowercase character i.e. “tutorial” with a single operation.

Input

str = “point”


Output

0


The string is already in the specified format.

There are two possible outcomes:

• The string's lowercase characters should all be converted to uppercase ones by determining the index of the last uppercase character.

• Alternately, get the index of the first lowercase character in the string and convert all the capital letters that come after it to lowercase.

Choose the situation where there are the fewest possible tasks.

## Java implementation

Let’s look into the Java implementation of this approach:

### Example

public class Main {
public static int minOperationsLower(String str, int n) {
// Store the indices of the last uppercase character and the first lowercase character
int i, lastUpperCase = -1, firstLowerCase = -1;

for (i = n - 1; i >= 0; i--) {  //finding uppercase
if (Character.isUpperCase(str.charAt(i))) {
lastUpperCase = i;
break;
}
}
for (i = 0; i < n; i++) {  //finding lowercase
if (Character.isLowerCase(str.charAt(i))) {
firstLowerCase = i;
break;
}
}
if (lastUpperCase == -1 || firstLowerCase == -1)
return 0;

// Counting of uppercase characters that appear after the first lowercase character
int countUpperCase = 0;
for (i = firstLowerCase; i < n; i++) {
if (Character.isUpperCase(str.charAt(i))) {
countUpperCase++;
}
}

// Count of lowercase characters that appear before the last uppercase character
int countLowerCase = 0;
for (i = 0; i < lastUpperCase; i++) {
if (Character.isLowerCase(str.charAt(i))) {
countLowerCase++;
}
}

// Return the minimum operations required
return Math.min(countLowerCase, countUpperCase);
}

// main method
public static void main(String args[])  {
String str = "tutoRIalsPoinT";
int n = str.length();
System.out.println("Operations Required: "+minOperationsLower(str, n));
}
}


### Output

Operations Required: 4


Time Complexity: O(N), where N is length of the string

Space Complexity: O(1), no extra space used.

Updated on: 22-Aug-2023

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