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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
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