Minimum Distance to the Target Element

Minimum Distance to the Target Element - Problem

Array Easy

Given an integer array nums (0-indexed) and two integers target and start, find an index i such that nums[i] == target and abs(i - start) is minimized.

Note: abs(x) is the absolute value of x.

Return abs(i - start).

It is guaranteed that target exists in nums.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4,5], target = 5, start = 3

› Output: 1

💡 Note: Target 5 is at index 4. Distance from start=3 is |4-3| = 1

Example 2 — Multiple Targets

$ Input: nums = [1], target = 1, start = 0

› Output: 0

💡 Note: Target 1 is at index 0, same as start. Distance is |0-0| = 0

Example 3 — Choose Closer Target

$ Input: nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0

› Output: 0

💡 Note: Target 1 appears everywhere, including at start position. Minimum distance is 0

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 10⁴
0 ≤ start < nums.length
target is in nums

Visualization

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Asked in

f Facebook 15 M Microsoft 12 🍎 Apple 8

The key insight is to iterate through the array once, checking each element against the target and calculating the distance from the start position. The optimal approach uses a single pass with early termination when distance 0 is found. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Occurrences

⏱️ Time: O(n) Space: O(1)

Iterate through the array to find all indices where nums[i] equals target, calculate the distance from each to start, and return the minimum distance.

Early Termination Optimization

⏱️ Time: O(n) Space: O(1)

Same approach as brute force, but with an optimization: if we find the target at the start position (distance = 0), we can immediately return since this is the minimum possible distance.

Brute Force - Check All Occurrences — Algorithm Steps

Find all indices where nums[i] == target
Calculate abs(i - start) for each index
Return the minimum distance

Visualization

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Step-by-Step Walkthrough

Scan Array

Check each element for target value

Calculate Distance

For matches, compute abs(i - start)

Track Minimum

Keep the smallest distance found

Code -

solution.c — C

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

int solution(int* nums, int numsSize, int target, int start) {
    int minDistance = INT_MAX;
    
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == target) {
            int distance = abs(i - start);
            if (distance < minDistance) {
                minDistance = distance;
            }
        }
    }
    
    return minDistance;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int target, start;
    scanf("%d", &target);
    scanf("%d", &start);
    
    int result = solution(nums, numsSize, target, start);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to check each element

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track minimum distance

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Occurrences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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