Kth Smallest Product of Two Sorted Arrays - Problem

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

long long countProducts(int* nums1, int nums1Size, int* nums2, int nums2Size, long long target) {
    long long count = 0;
    for (int i = 0; i < nums1Size; i++) {
        int num1 = nums1[i];
        if (num1 == 0) {
            if (target >= 0) count += nums2Size;
            continue;
        }
        
        if (num1 > 0) {
            int left = 0, right = nums2Size - 1;
            while (left <= right) {
                int mid = (left + right) / 2;
                if ((long long)num1 * nums2[mid] <= target) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
            count += left;
        } else {
            int left = 0, right = nums2Size - 1;
            while (left <= right) {
                int mid = (left + right) / 2;
                if ((long long)num1 * nums2[mid] <= target) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            count += nums2Size - right - 1;
        }
    }
    return count;
}

long long solution(int* nums1, int nums1Size, int* nums2, int nums2Size, int k) {
    // Find bounds - check all corner products
    long long left = LLONG_MAX, right = LLONG_MIN;
    
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
            int idx1 = (i == 0) ? 0 : nums1Size - 1;
            int idx2 = (j == 0) ? 0 : nums2Size - 1;
            long long prod = (long long)nums1[idx1] * nums2[idx2];
            if (prod < left) left = prod;
            if (prod > right) right = prod;
        }
    }
    
    while (left < right) {
        long long mid = left + (right - left) / 2;
        if (countProducts(nums1, nums1Size, nums2, nums2Size, mid) >= k) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    
    return left;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    
    // Skip to opening bracket
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        // Skip whitespace and commas
        while (*p == ' ' || *p == ',' || *p == '\t' || *p == '\n') p++;
        
        if (*p == ']' || *p == '\0') break;
        
        // Parse number using strtol
        char* endptr;
        long val = strtol(p, &endptr, 10);
        if (endptr != p) {  // Successfully parsed a number
            arr[(*size)++] = (int)val;
            p = endptr;
        } else {
            break;
        }
    }
}

int main() {
    char line[10000];
    int nums1[1000], nums1Size;
    int nums2[1000], nums2Size;
    
    // Parse nums1
    if (fgets(line, sizeof(line), stdin)) {
        parseArray(line, nums1, &nums1Size);
    }
    
    // Parse nums2
    if (fgets(line, sizeof(line), stdin)) {
        parseArray(line, nums2, &nums2Size);
    }
    
    // Parse k
    int k = 0;
    if (fgets(line, sizeof(line), stdin)) {
        k = atoi(line);
    }
    
    printf("%lld\n", solution(nums1, nums1Size, nums2, nums2Size, k));
    return 0;
}