Score of a String - Practice Coding Problems

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.

For example, if we have the string "hello", we calculate:

|ASCII('h') - ASCII('e')| = |104 - 101| = 3
|ASCII('e') - ASCII('l')| = |101 - 108| = 7
|ASCII('l') - ASCII('l')| = |108 - 108| = 0
|ASCII('l') - ASCII('o')| = |108 - 111| = 3

The total score would be 3 + 7 + 0 + 3 = 13.

Goal: Return the score of the given string s.

Input & Output

example_1.py — Basic Example

$ Input: s = "hello"

› Output: 13

💡 Note: The adjacent character differences are: |'h'-'e'| = |104-101| = 3, |'e'-'l'| = |101-108| = 7, |'l'-'l'| = |108-108| = 0, |'l'-'o'| = |108-111| = 3. Total: 3+7+0+3 = 13

example_2.py — Single Character

$ Input: s = "z"

› Output: 0

💡 Note: A single character has no adjacent characters, so the score is 0

example_3.py — Same Characters

$ Input: s = "aaa"

› Output: 0

💡 Note: All characters are the same, so all adjacent differences are |'a'-'a'| = 0. Total: 0+0 = 0

Constraints

2 ≤ s.length ≤ 100
s consists only of lowercase English letters
ASCII values of lowercase letters range from 97 ('a') to 122 ('z')

Visualization

Tap to expand

Understanding the Visualization

Map Characters to Elevations

Each character becomes a checkpoint with ASCII value as elevation

Calculate Elevation Changes

Find absolute difference between consecutive checkpoint elevations

Sum Total Effort

Add all elevation changes to get total hiking effort (score)

Key Takeaway

🎯 Key Insight: We only need to visit each character once and calculate the ASCII difference with its immediate neighbor, making this an optimal O(n) solution.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 10

The optimal solution uses a single pass through the string, calculating the absolute difference between each pair of adjacent characters. With O(n) time complexity and O(1) space complexity, this approach efficiently computes the string score by iterating once and summing the ASCII value differences.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Check all possible character pairs in the string
Single Pass (Optimal)	O(n)	O(1)	Iterate through string once, calculating adjacent differences

Brute Force (Nested Loops) — Algorithm Steps

Iterate through each character with outer loop
For each character, iterate through all other characters with inner loop
Calculate absolute difference for adjacent pairs only
Sum up all the differences

Visualization

Tap to expand

Step-by-Step Walkthrough

Check All Pairs

Outer loop at position i, inner loop checks all positions j

Filter Adjacent

Only calculate difference when j = i + 1

Sum Differences

Add absolute difference to running total

Code -

solution.c — C

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

int scoreOfString(char* s) {
    int score = 0;
    int n = strlen(s);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            // Only consider adjacent pairs
            if (j == i + 1) {
                score += abs(s[i] - s[j]);
            }
        }
    }
    return score;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    // Remove newline and quotes
    s[strcspn(s, "\n")] = 0;
    if (s[0] == '"' && s[strlen(s)-1] == '"') {
        memmove(s, s+1, strlen(s));
        s[strlen(s)-1] = '\0';
    }
    printf("%d\n", scoreOfString(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops iterate through all character pairs, even though we only need adjacent ones

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to store the running sum

✓ Linear Space

58.0K Views

Medium Frequency

~5 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler