Given a string str, containing upper case and lower case characters. In a single operations, any lowercase character can be converted to an uppercase character and vice versa. The task is to print the minimum number of such operations required so that the resultant string consists of zero or more upper case characters followed by zero or more lower case characters.
Examples:
Input: str = “geEks”
Output: 1
Either the first 2 characters can be converted to uppercase characters i.e. “GEEks” with 2 operations.
Or the third character can be converted to a lowercase character i.e. “geeks” with a single operation.Input: str = “geek”
Output: 0
The string is already in the specified format.
Approach:
There are two possible cases:
- Find the index of the last uppercase character in the string and convert all the lowercase characters appearing before it into uppercase characters.
- Or, find the index of the first lowercase character in the string and convert all the uppercase characters appearing after it into lowercase characters.
Choose the case where the operations required are minimum.
Below is the implementation of the above approach:
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std;
// Function to return the minimum // number of operations required int minOperations(string str, int n)
{ // To store the indices of the last uppercase
// and the first lowercase character
int i, lastUpper = -1, firstLower = -1;
// Find the last uppercase character
for (i = n - 1; i >= 0; i--)
{
if ( isupper (str[i]))
{
lastUpper = i;
break ;
}
}
// Find the first lowercase character
for (i = 0; i < n; i++)
{
if ( islower (str[i]))
{
firstLower = i;
break ;
}
}
// If all the characters are either
// uppercase or lowercase
if (lastUpper == -1 || firstLower == -1)
return 0;
// Count of uppercase characters that appear
// after the first lowercase character
int countUpper = 0;
for (i = firstLower; i < n; i++)
{
if ( isupper (str[i]))
{
countUpper++;
}
}
// Count of lowercase characters that appear
// before the last uppercase character
int countLower = 0;
for (i = 0; i < lastUpper; i++)
{
if ( islower (str[i]))
{
countLower++;
}
}
// Return the minimum operations required
return min(countLower, countUpper);
} // Driver Code int main()
{ string str = "geEksFOrGEekS" ;
int n = str.length();
cout << minOperations(str, n) << endl;
} // This code is contributed by // Surendra_Gangwar |
// Java implementation of the approach class GFG {
// Function to return the minimum
// number of operations required
static int minOperations(String str, int n)
{
// To store the indices of the last uppercase
// and the first lowercase character
int i, lastUpper = - 1 , firstLower = - 1 ;
// Find the last uppercase character
for (i = n - 1 ; i >= 0 ; i--) {
if (Character.isUpperCase(str.charAt(i))) {
lastUpper = i;
break ;
}
}
// Find the first lowercase character
for (i = 0 ; i < n; i++) {
if (Character.isLowerCase(str.charAt(i))) {
firstLower = i;
break ;
}
}
// If all the characters are either
// uppercase or lowercase
if (lastUpper == - 1 || firstLower == - 1 )
return 0 ;
// Count of uppercase characters that appear
// after the first lowercase character
int countUpper = 0 ;
for (i = firstLower; i < n; i++) {
if (Character.isUpperCase(str.charAt(i))) {
countUpper++;
}
}
// Count of lowercase characters that appear
// before the last uppercase character
int countLower = 0 ;
for (i = 0 ; i < lastUpper; i++) {
if (Character.isLowerCase(str.charAt(i))) {
countLower++;
}
}
// Return the minimum operations required
return Math.min(countLower, countUpper);
}
// Driver Code
public static void main(String args[])
{
String str = "geEksFOrGEekS" ;
int n = str.length();
System.out.println(minOperations(str, n));
}
} |
# Python 3 implementation of the approach # Function to return the minimum # number of operations required def minOperations( str , n):
# To store the indices of the last uppercase
# and the first lowercase character
lastUpper = - 1
firstLower = - 1
# Find the last uppercase character
for i in range ( n - 1 , - 1 , - 1 ):
if ( str [i].isupper()):
lastUpper = i
break
# Find the first lowercase character
for i in range (n):
if ( str [i].islower()):
firstLower = i
break
# If all the characters are either
# uppercase or lowercase
if (lastUpper = = - 1 or firstLower = = - 1 ):
return 0
# Count of uppercase characters that appear
# after the first lowercase character
countUpper = 0
for i in range ( firstLower,n):
if ( str [i].isupper()):
countUpper + = 1
# Count of lowercase characters that appear
# before the last uppercase character
countLower = 0
for i in range (lastUpper):
if ( str [i].islower()):
countLower + = 1
# Return the minimum operations required
return min (countLower, countUpper)
# Driver Code if __name__ = = "__main__" :
str = "geEksFOrGEekS"
n = len ( str )
print (minOperations( str , n))
# This code is contributed by Ita_c. |
// C# implementation of the approach using System;
class GFG
{ // Function to return the minimum
// number of operations required
static int minOperations( string str, int n)
{
// To store the indices of the last uppercase
// and the first lowercase character
int i, lastUpper = -1, firstLower = -1;
// Find the last uppercase character
for (i = n - 1; i >= 0; i--)
{
if (Char.IsUpper(str[i]))
{
lastUpper = i;
break ;
}
}
// Find the first lowercase character
for (i = 0; i < n; i++)
{
if (Char.IsLower(str[i]))
{
firstLower = i;
break ;
}
}
// If all the characters are either
// uppercase or lowercase
if (lastUpper == -1 || firstLower == -1)
return 0;
// Count of uppercase characters that appear
// after the first lowercase character
int countUpper = 0;
for (i = firstLower; i < n; i++)
{
if (Char.IsUpper(str[i]))
{
countUpper++;
}
}
// Count of lowercase characters that appear
// before the last uppercase character
int countLower = 0;
for (i = 0; i < lastUpper; i++)
{
if (Char.IsLower(str[i]))
{
countLower++;
}
}
// Return the minimum operations required
return Math.Min(countLower, countUpper);
}
// Driver Code
public static void Main()
{
string str = "geEksFOrGEekS" ;
int n = str.Length;
Console.WriteLine(minOperations(str, n));
}
} // This code is contributed by Ryuga |
<?php // PHP implementation of the approach // Function to return the minimum // number of operations required function minOperations( $str , $n )
{ // To store the indices of the last uppercase
// and the first lowercase character
$i ; $lastUpper = -1; $firstLower = -1;
// Find the last uppercase character
for ( $i = $n - 1; $i >= 0; $i --)
{
if (ctype_upper( $str [ $i ]))
{
$lastUpper = $i ;
break ;
}
}
// Find the first lowercase character
for ( $i = 0; $i < $n ; $i ++)
{
if (ctype_lower( $str [ $i ]))
{
$firstLower = $i ;
break ;
}
}
// If all the characters are either
// uppercase or lowercase
if ( $lastUpper == -1 || $firstLower == -1)
return 0;
// Count of uppercase characters that appear
// after the first lowercase character
$countUpper = 0;
for ( $i = $firstLower ; $i < $n ; $i ++)
{
if (ctype_upper( $str [ $i ]))
{
$countUpper ++;
}
}
// Count of lowercase characters that appear
// before the last uppercase character
$countLower = 0;
for ( $i = 0; $i < $lastUpper ; $i ++)
{
if (ctype_lower( $str [ $i ]))
{
$countLower ++;
}
}
// Return the minimum operations required
return min( $countLower , $countUpper );
}
// Driver Code
{
$str = "geEksFOrGEekS" ;
$n = strlen ( $str );
echo (minOperations( $str , $n ));
}
// This code is contributed by Code_Mech ?> |
<script> // JavaScript implementation of the approach
//Check UpperCase
function isupper(str) {
return str === str.toUpperCase();
}
//Check UpperCase
function islower(str) {
return str === str.toLowerCase();
}
// Function to return the minimum
// number of operations required
function minOperations(str, n) {
// To store the indices of the last uppercase
// and the first lowercase character
var i,
lastUpper = -1,
firstLower = -1;
// Find the last uppercase character
for (i = n - 1; i >= 0; i--) {
if (isupper(str[i])) {
lastUpper = i;
break ;
}
}
// Find the first lowercase character
for (i = 0; i < n; i++) {
if (islower(str[i])) {
firstLower = i;
break ;
}
}
// If all the characters are either
// uppercase or lowercase
if (lastUpper === -1 || firstLower === -1) return 0;
// Count of uppercase characters that appear
// after the first lowercase character
var countUpper = 0;
for (i = firstLower; i < n; i++) {
if (isupper(str[i])) {
countUpper++;
}
}
// Count of lowercase characters that appear
// before the last uppercase character
var countLower = 0;
for (i = 0; i < lastUpper; i++) {
if (islower(str[i])) {
countLower++;
}
}
// Return the minimum operations required
return Math.min(countLower, countUpper);
}
// Driver Code
var str = "geEksFOrGEekS" ;
var n = str.length;
document.write(minOperations(str, n) + "<br>" );
</script>
|
6
Complexity Analysis:
- Time Complexity: O(N) where N is the length of the string.
- Auxiliary Space: O(1), since no extra space has been taken.