LeetCode #1848 — EASY

Minimum Distance to the Target Element

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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 that abs(x) is the absolute value of x.

Return abs(i - start).

It is guaranteed that target exists in nums.

Example 1:

Input: nums = [1,2,3,4,5], target = 5, start = 3
Output: 1
Explanation: nums[4] = 5 is the only value equal to target, so the answer is abs(4 - 3) = 1.

Example 2:

Input: nums = [1], target = 1, start = 0
Output: 0
Explanation: nums[0] = 1 is the only value equal to target, so the answer is abs(0 - 0) = 0.

Example 3:

Input: nums = [1,1,1,1,1,1,1,1,1,1], target = 1, start = 0
Output: 0
Explanation: Every value of nums is 1, but nums[0] minimizes abs(i - start), which is abs(0 - 0) = 0.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 that abs(x) is the absolute value of x. Return abs(i - start). It is guaranteed that target exists in nums.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[1,2,3,4,5]
5
3

Example 2

[1]
1
0

Example 3

[1,1,1,1,1,1,1,1,1,1]
1
0

Step 02

Core Insight

What unlocks the optimal approach

Loop in both directions until you find the target element.
For each index i such that nums[i] == target calculate abs(i - start).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1848: Minimum Distance to the Target Element
class Solution {
    public int getMinDistance(int[] nums, int target, int start) {
        int n = nums.length;
        int ans = n;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == target) {
                ans = Math.min(ans, Math.abs(i - start));
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1848: Minimum Distance to the Target Element
func getMinDistance(nums []int, target int, start int) int {
	ans := 1 << 30
	for i, x := range nums {
		if t := abs(i - start); x == target && t < ans {
			ans = t
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1848: Minimum Distance to the Target Element
class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        return min(abs(i - start) for i, x in enumerate(nums) if x == target)

// Accepted solution for LeetCode #1848: Minimum Distance to the Target Element
impl Solution {
    pub fn get_min_distance(nums: Vec<i32>, target: i32, start: i32) -> i32 {
        nums.iter()
            .enumerate()
            .filter(|&(_, &x)| x == target)
            .map(|(i, _)| ((i as i32) - start).abs())
            .min()
            .unwrap_or_default()
    }
}

// Accepted solution for LeetCode #1848: Minimum Distance to the Target Element
function getMinDistance(nums: number[], target: number, start: number): number {
    let ans = Infinity;
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i] === target) {
            ans = Math.min(ans, Math.abs(i - start));
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.