Find the Distance Value Between Two Arrays

Find the Distance Value Between Two Arrays - Problem

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i] - arr2[j]| <= d.

In other words, count how many elements in arr1 are more than distance d away from all elements in arr2.

Input & Output

Example 1 — Basic Case

$ Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2

› Output: 2

💡 Note: For arr1[0]=4: closest in arr2 is 1, |4-1|=3 > 2 ✓. For arr1[1]=5: closest is 1, |5-1|=4 > 2 ✓. For arr1[2]=8: found 8 in arr2, |8-8|=0 ≤ 2 ✗. Result: 2 elements count.

Example 2 — All Elements Count

$ Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3

› Output: 2

💡 Note: For arr1[0]=1: closest in arr2 is -3, |1-(-3)|=4 > 3 ✓. For arr1[1]=4: closest is 6, |4-6|=2 ≤ 3 ✗. For arr1[2]=2: closest is -3, |2-(-3)|=5 > 3 ✓. For arr1[3]=3: closest is 6, |3-6|=3 ≤ 3 ✗. Result: 2 elements count.

Example 3 — No Elements Count

$ Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6

› Output: 1

💡 Note: Only arr1[2]=100 is far enough from all arr2 elements. The others have at least one arr2 element within distance 6.

Constraints

1 ≤ arr1.length, arr2.length ≤ 500
-1000 ≤ arr1[i], arr2[j] ≤ 1000
0 ≤ d ≤ 100

Visualization

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Asked in

f Facebook 15 M Microsoft 12 a Amazon 8

The key insight is that for each element in arr1, we need to check if ALL elements in arr2 are more than distance d away. Best approach uses sorting + binary search to efficiently find the closest elements. Time: O(m log m + n log m), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n×m) Space: O(1)

For every element in arr1, iterate through all elements in arr2 to see if any element is within distance d. If no element in arr2 is within distance d, increment the counter.

Sort Then Search

⏱️ Time: O(m log m + n log m) Space: O(1)

Sort arr2 to enable efficient searching. For each element in arr1, use binary search to find the closest elements in arr2 and check if any are within distance d.

Brute Force — Algorithm Steps

For each element arr1[i]
Check all elements arr2[j]
If |arr1[i] - arr2[j]| <= d for any j, skip this element
Otherwise increment count

Visualization

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Step-by-Step Walkthrough

Pick Element

Take first element from arr1

Check All

Compare against every element in arr2

Count Valid

If no element is within distance d, increment counter

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int bisectLeft(int* arr, int size, int x) {
    int left = 0;
    int right = size;
    while (left < right) {
        int mid = (left + right) / 2;
        if (arr[mid] < x) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
}

int myabs(int x) {
    return x < 0 ? -x : x;
}

int solution(int* arr1, int arr1Size, int* arr2, int arr2Size, int d) {
    qsort(arr2, arr2Size, sizeof(int), compare);
    int count = 0;
    
    for (int i = 0; i < arr1Size; i++) {
        int num1 = arr1[i];
        // Find insertion point
        int pos = bisectLeft(arr2, arr2Size, num1);
        int isValid = 1;
        
        // Check element at pos (if exists)
        if (pos < arr2Size && myabs(num1 - arr2[pos]) <= d) {
            isValid = 0;
        }
        
        // Check element before pos (if exists)
        if (isValid && pos > 0 && myabs(num1 - arr2[pos - 1]) <= d) {
            isValid = 0;
        }
        
        if (isValid) {
            count++;
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    char* endptr;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, &endptr, 10);
        p = endptr;
    }
}

int main() {
    char line[1000];
    
    // Read first array
    fgets(line, sizeof(line), stdin);
    int arr1[1000];
    int arr1Size;
    parseArray(line, arr1, &arr1Size);
    
    // Read second array
    fgets(line, sizeof(line), stdin);
    int arr2[1000];
    int arr2Size;
    parseArray(line, arr2, &arr2Size);
    
    // Read d
    int d;
    scanf("%d", &d);
    
    int result = solution(arr1, arr1Size, arr2, arr2Size, d);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

For each of n elements in arr1, we check all m elements in arr2

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Common Approaches

Brute Force — Algorithm Steps

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Code -

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