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: 52

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

Approach	Time	Space	Notes
✓ Single Pass Iteration	O(n)	O(1)	Iterate through string and sum absolute differences of adjacent characters
Optimized Single Pass with Early Termination	O(n)	O(1)	Same algorithm but with potential optimizations for special cases

Single Pass Iteration — Algorithm Steps

Initialize sum to 0
Loop from index 1 to string length
For each position, calculate |ASCII(s[i]) - ASCII(s[i-1])|
Add difference to sum
Return total sum

Visualization

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Step-by-Step Walkthrough

Initialize

Start with score = 0

Compare Adjacent

Calculate |ASCII(current) - ASCII(previous)|

Sum Up

Add each difference to running total

Code -

solution.c — C

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

int solution(char* s) {
    int score = 0;
    int len = strlen(s);
    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

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Single Pass Iteration — Algorithm Steps

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Code -

Time & Space Complexity

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