Pattern – Integer to Alphabets

Convert a given integer to its corresponding alphabetic representation where 1=A, 2=B, ..., 26=Z, 27=AA, 28=AB, etc.

Understand the Problem

Problem Statement

Given a number N, convert the number to character values based on alphabet positions. The mapping follows a pattern where 1 maps to 'A', 2 to 'B', ..., 26 to 'Z', 27 to 'AA', 28 to 'AB', and so on. This is similar to how Excel column labels are generated.

Constraints

1 ≤ N ≤ 10,000,000
N is a positive integer
Output should be uppercase alphabetic characters only

Examples

Example 1

Input

Output

Explanation

10 maps directly to the 10th letter of the alphabet, which is 'J'.

Example 2

Input

Output

AB

Explanation

28 = 26 + 2, so it maps to the 2nd letter 'B' with one complete cycle of 26, giving us 'AB' (where A represents one complete cycle of 26).

Solution

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

int main() {
    long long n;
    scanf("%lld", &n);
    
    char result[100];
    int index = 0;
    
    if (n <= 26) {
        printf("%c", 'A' + n - 1);
        return 0;
    }
    
    while (n > 26) {
        int remainder = n % 26;
        if (remainder == 0) {
            result[index++] = 'Z';
            n = n / 26 - 1;
        } else {
            result[index++] = 'A' + remainder - 1;
            n = n / 26;
        }
    }
    
    result[index++] = 'A' + n - 1;
    
    for (int i = index - 1; i >= 0; i--) {
        printf("%c", result[i]);
    }
    
    return 0;
}

Time:O(log N)

Space:O(log N)

Approach:

The C solution follows the same algorithmic approach. We use a character array to store the result characters. The key insight is handling the case when n % 26 equals 0 - this represents a 'Z' in the current position, and we need to reduce the quotient by 1 to account for the fact that we don't have a zero in our alphabet system. After building the string from right to left, we print it in reverse order.

Visual Explanation

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