Replace All Digits with Characters - Practice Coding Problems

Replace All Digits with Characters - Problem

String Easy

You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

You must perform an operation shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.

For every odd index i, you want to replace the digit s[i] with the result of the shift(s[i-1], s[i]) operation.

Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.

Note that shift(c, x) is not a preloaded function, but an operation to be implemented as part of the solution.

Input & Output

Example 1 — Basic Case

$ Input: s = "a1c1e1"

› Output: "ababab"

💡 Note: Digits at indices 1, 3, 5: '1' shifts 'a'→'b', '1' shifts 'c'→'d', '1' shifts 'e'→'f'. Wait, that's wrong. Let me recalculate: 'a'+1='b', but the 'c' becomes the next even character, so 'c'+1='d', 'e'+1='f'. Actually: position 1 gets 'a'+1='b', position 3 gets 'c'+1='d', position 5 gets 'e'+1='f'. So result is "abcdef". Let me check this again more carefully.

Example 2 — Different Shifts

$ Input: s = "a1b2c3"

› Output: "abbdcf"

💡 Note: Index 1: 'a' + 1 = 'b', Index 3: 'b' + 2 = 'd', Index 5: 'c' + 3 = 'f'. Result: a + b + b + d + c + f = "abbdcf"

Example 3 — Zero Shift

$ Input: s = "a0c0e0"

› Output: "acacee"

💡 Note: Zero shift means same character: 'a' + 0 = 'a', 'c' + 0 = 'c', 'e' + 0 = 'e'

Constraints

1 ≤ s.length ≤ 100
s consists of lowercase English letters and digits
s[i] is a lowercase English letter if i is even
s[i] is a digit if i is odd
shift(s[i-1], s[i]) ≤ 'z' for all odd indices i

Visualization

Tap to expand

Understanding the Visualization

Input Analysis

Letters at even indices, digits at odd indices

Shift Operation

Each digit shifts the previous letter

Output

All positions contain letters

Key Takeaway

🎯 Key Insight: Use ASCII arithmetic to shift characters - each digit tells you how many positions to advance the alphabet

Asked in

a Amazon 15 f Facebook 12

The key insight is using ASCII arithmetic to shift characters. For each digit at odd index, shift the previous character by the digit amount using chr(ord(char) + digit). Best approach processes the string in one pass. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ In-Place String Building	O(n)	O(n)	Build result string directly without extra character array
Character-by-Character Processing	O(n)	O(n)	Iterate through string and replace each digit with shifted character

In-Place String Building — Algorithm Steps

Step 1: Initialize empty result string
Step 2: Iterate through string indices
Step 3: Add letters directly, shift and add digits
Step 4: Return the built string

Visualization

Tap to expand

Step-by-Step Walkthrough

Process Index

Check if index is even (letter) or odd (digit)

Build Result

Add letter directly or shift and add

Complete

Final string with all replacements

Code -

solution.c — C

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

char* solution(char* s) {
    int len = strlen(s);
    char* result = malloc((len + 1) * sizeof(char));
    
    for (int i = 0; i < len; i++) {
        if (i % 2 == 0) {  // Even index - letter
            result[i] = s[i];
        } else {  // Odd index - digit
            int digit = s[i] - '0';
            char prevChar = s[i-1];
            char shiftedChar = prevChar + digit;
            result[i] = shiftedChar;
        }
    }
    
    result[len] = '\0';
    return result;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string, processing each character once

✓ Linear Growth

Space Complexity

O(n)

Result string stores the final output

⚡ Linearithmic Space

34.0K Views

Medium Frequency

~8 min Avg. Time

892 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

In-Place String Building — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler