Algorithm Logic

The task requires implementing a function that calculates the frequency of a specific digit (0-9) within a given integer. The primary challenge involves handling the sign of the input number and the edge case where the input is zero.

Since the input integer is passed as a const int, it cannot be modified directly. Therefore, a local variable is used to work with the absolute value of the number. A standard loop iterates through the digits by using the modulo operator to extract the least significant digit and integer division to reduce the number.

Edge Case Handling

A specific condition arises when the input integer is 0 and the target digit is also 0. In this scenario, the loop condition typically used for positive integers would fail to execute, returning 0 instead of 1. An explicit check is required to return 1 in this instance.

Implementation 1: Direct Iteration

int countDigitOccurrences(const int number, const int target) {
    long currentVal = number;
    if (currentVal < 0) {
        currentVal = -currentVal;
    }

    if (target == 0 && number == 0) {
        return 1;
    }

    int frequency = 0;
    while (currentVal > 0) {
        int digit = currentVal % 10;
        if (digit == target) {
            frequency++;
        }
        currentVal /= 10;
    }
    return frequency;
}

Implementation 2: Histogram Approach

An alternative strategy involves utilizing an array of size 10 to record the count of every digit (0-9) present in the number. This allows for O(1) lookup after the iteration completes.

int countDigitHistogram(const int number, const int target) {
    int absoluteVal = (number < 0) ? -number : number;
    int buckets[10] = {0};

    if (absoluteVal == 0) {
        buckets[0] = 1;
    } else {
        while (absoluteVal > 0) {
            int digit = absoluteVal % 10;
            buckets[digit]++;
            absoluteVal /= 10;
        }
    }
    return buckets[target];
}