Number of Larger Elements on right side in a string in C++

Given a string, we have to count the number of larger elements on right side of each character. Let's see an example.

Input

string = "abc"

Output

2 1 0

There are 2 larger elements than a on its right side.

There is 1 larger element than b on its right side.

There are 0 larger elements than c on its right side.

Algorithm

• Initialise the string.

• Initialise an array to keep track of the count.

• Write two loops to iterate over the string.

  • Take one char at a time and compare it with all the character after it.

  • Increment the corresponding character count in the count array if the current element is less than the next element.

• Print the count of all characters.

Implementation

Following is the implementation of the above algorithm in C++

Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
void countCharNextLargerElementsCount(string str) {
   int len = str.length(), count[len];
   for (int i = 0; i < len; i++) {
      count[i] = 0;
   }
   for (int i = 0; i < len; i++) {
      for (int j = i + 1; j < len; j++) {
         if (str[i] < str[j]) {
            count[i]++;
         }
      }
   }
   for (int i = 0; i < len; i++) {
      cout << count[i] << " ";
   }
   cout << endl;
}
int main() {
   string str = "abcdefgh";
   countCharNextLargerElementsCount(str);
   return 0;
}


Output


If you run the above code, then you will get the following result.
7 6 5 4 3 2 1 0