# Sum of Indices of Characters Removed to Obtain an Empty String Based On Given Conditions

String manipulation-related concepts like The Sum of indices of Characters deleted to obtain an Empty String are frequently utilized in programming challenges and competitions. The result is then computed using the sum of the eliminated characters' indexes.

The Sum of indices of Characters deleted to obtain an Empty String is a practical idea in string manipulation that may be used to solve a variety of programming difficulties and challenges.

## Problem Approach

We must first comprehend the problem statement and given criteria to find total of character indices eliminated to produce empty string.

Given string S, the objective is to determine the total number of characters that can be removed from S while still leaving string that is empty. For instance, if S = "code," characters at positions 0, 4, 5, and 6 can be eliminated to get empty string. These indices add up to 0 + 4 + 5 + 6 = 15.

However, using stack is a typical strategy for fixing this issue. We can loop through character string S and determine if each character can be deleted for each iteration. We can add its index to stack if it can be removed. If it is impossible to remove it, we can see if the character at top of stack can be eliminated along with one now in place. If it can be eliminated, we will do so and add its index to sum together with index of current character. This procedure can be repeated until all the characters in string have been handled.

The following pseudocode exemplifies this strategy −

stack = []
sum = 0
for k in range(len(S)):
if stack and S[k] == S[stack[-1]]:
stack.pop()
sum += k + stack[-1] if stack else k
else:
stack.append(k)
return sum


In this pseudocode, sum variable and an empty stack are both initialized to 0. A for loop is then used to repeatedly iterate through string S. Each character is examined to see if it may be deleted along with the character at top of stack and if stack is not empty. The character is removed from stack if it can be done so and sum of its index and character in play is added to sum variable. In that case, we add its index to the stack and attempt to remove it. We then give back the sum variable.

The time and space complexity of this method are both O(n), where n is length of string S and n is maximum number of characters that can be dropped from S.

### Syntax

The C++ syntax to determine total number of character indices eliminated to create empty string based on specified criteria is as follows −

### Explanation

• We start by getting the user’s input for string.

• We set n's starting value to be string's length, str.

• Next, we initialize cnt to 0, which will count occurrences of character "U".

• We set the initial value of sum to 0, which will store total number of eliminated character indices.

• After that, we loop through str, checking each character as follows −

• If character is "U," we raise cnt and increase sum by (n - i - 1) + 2 * cnt.

• If character is not "U," we increase sum by adding i + 2 * cnt.

• Lastly, we output sum's value.

Note − Because specified circumstances for this problem are not stated clearly in question, presumptively used those conditions.

{
string str;
cin >> str;

int n = str.size();
int cnt = 0, sum = 0;
for (int k = 0; i < n; k++) {
if (str[k] == 'U') {
sum += (n - k - 1) + 2 * cnt;
cnt++;
} else {
sum += k + 2 * cnt;
}
}
cout << sum << endl;
}


### Algorithm

A C++ algorithm to calculate total number of character indices eliminated to create an empty string under defined conditions −

• Step 1 − To begin, define a string variable and input user-supplied string.

• Step 2 − Make a stack to hold string's characters.

• Step 3 − The input string is looped through character by character.

• Step 4 − Push current character into stack if it is empty.

• Step 5 − Pop top character from the stack if current character and top character of stack are identical.

• Step 6 − Push current character into stack if it differs from character at top of stack.

• Step 7 − Only characters that cannot be erased will remain in stack following loop.

• Step 8 − Add up indices of characters that are still in stack.

• Step 9 − Display total of indices.

## Approaches to Follow

### Approach-1

Using following condition to computes sum of character removal indexes to produce empty string −

The string "abacbdc" is used as input in this example. The two indices i and j, are used by code to traverse string from start to finish. Conditions for character removal from string are as follows:

If s[i] and s[j] are equal, shift both indices to center of string.

• Remove character at index j and increase sum of indices by index i+1 if s[i] is smaller than s[j].

• Remove character at index i and increase sum of indices by index j+1 if s[i] is greater than s[j].

The total of indices is then reported to console after all characters have been eliminated.

Remember that this is merely illustration and that requirements for character removal may change depending on the nature of the issue.

### Example-1

#include <iostream>
#include <string>

using namespace std;

int main() {
string s = "abacbdc";
int sum = 0;
int i = 0;
int j = s.length() - 1;
while (i < j) {
if (s[i] == s[j]) {
i++;
j--;
} else if (s[i] < s[j]) {
sum += i + 1;
i++;
s.erase(j, 1);
j--;
} else {
sum += j + 1;
j--;
s.erase(i, 1);
i++;
}
}
cout << "Sum of indices of characters removed: " << sum << endl;
return 0;
}


### Output

Sum of indices of characters removed: 6


### Approach-2

The str string and character are inputs for sum_of_indices function. Then, iterating through string, it determines whether each character equals c. If so, function decrements loop index to account for removed character and adds character's index to running total before erasing character from string using erase technique. The function then returns the total number of eliminated character indices.

The sample string str and character c are defined in the main function, and these two inputs are used to invoke sum_of_indices function. The total is printed to console as result.

### Example-2

#include <iostream>
#include <string>
using namespace std;
int sum_of_indices(string str, char c) {
int sum = 0;
for (int i = 0; i < str.length(); i++) {
if (str[i] == c) {
sum += i;
str.erase(i, 1);
i--;
}
}
return sum;
}
int main() {
string str = "abcbcdc";
char c = 'c';
int sum = sum_of_indices(str, c);
cout << "Sum of indices of characters removed to obtain empty string: " << sum << endl;
return 0;
}


### Output

Sum of indices of characters removed to obtain empty string: 9


## Conclusion

Manipulating strings and their indices is required to solve the problem of calculating sum of indices of characters eliminated to obtain empty string based on supplied conditions. To solve this issue, cycle through string and, if two consecutive characters are same, delete them before updating index. We can add indices of characters that were eliminated to produce an empty string once we have one.

There are number of solutions to issue, such as utilizing stack or queue to keep track of characters to remove or recursion to iteratively remove characters from string.

Updated on: 10-May-2023

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