Score of a String

Score of a String - Problem

String Easy

You are given a string s. The score of a string is defined as the sum of the absolute difference between the ASCII values of adjacent characters.

Return the score of s.

Input & Output

Example 1 — Basic Case

$ Input: s = "hello"

› Output: 13

💡 Note: Adjacent differences: |'h'-'e'| = |104-101| = 3, |'e'-'l'| = |101-108| = 7, |'l'-'l'| = |108-108| = 0, |'l'-'o'| = |108-111| = 3. Sum = 3+7+0+3 = 13

Example 2 — Two Characters

$ Input: s = "zab"

› Output: 26

💡 Note: |'z'-'a'| = |122-97| = 25, |'a'-'b'| = |97-98| = 1. Sum = 25+1 = 26

Example 3 — Single Character

$ Input: s = "a"

› Output: 0

💡 Note: Single character has no adjacent pairs, so score is 0

Constraints

1 ≤ s.length ≤ 100
s consists of lowercase English letters only

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that we need to sum absolute differences between adjacent character ASCII values. The optimal approach iterates through the string once, calculating differences on the fly. Time: O(n), Space: O(1)

Common Approaches

✓ Optimized Single Pass with Early Termination

⏱️ Time: O(n) Space: O(1)

Uses the same single-pass approach but can include optimizations like handling empty strings or single characters without entering the loop.

Single Pass Iteration

⏱️ Time: O(n) Space: O(1)

Loop through the string from index 1 to end, calculate absolute difference between current and previous character's ASCII values, and sum them up.

Optimized Single Pass with Early Termination — Algorithm Steps

Check for edge cases (length < 2)
Initialize sum to 0
Loop through string calculating differences
Return sum

Visualization

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Step-by-Step Walkthrough

Edge Check

Return 0 if string length < 2

Calculate

Sum adjacent ASCII differences

Result

Return total score

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s) {
    int len = strlen(s);
    if (len < 2) {
        return 0;
    }
    
    int score = 0;
    for (int i = 1; i < len; i++) {
        score += abs(s[i] - s[i-1]);
    }
    return score;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using a single variable to store the sum

✓ Linear Space

12.0K Views

Medium Frequency

~5 min Avg. Time

450 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Single Pass with Early Termination — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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