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Formatted question description: https://leetcode.ca/all/1848.html

1848. Minimum Distance to the Target Element

Level

Easy

Description

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] <= 10^4
  • 0 <= start < nums.length
  • target is in nums.

Solution

Loop over nums. For each i such that nums[i] == target, calculate Math.abs(i - start) and maintain the minimum distance. Finally, return the minimum distance.

  • class Solution {
    public int getMinDistance(int[] nums, int target, int start) {
        int minDistance = Integer.MAX_VALUE;
        int length = nums.length;
        for (int i = 0; i < length; i++) {
            if (nums[i] == target) {
                int distance = Math.abs(i - start);
                minDistance = Math.min(minDistance, distance);
            }
        }
        return minDistance;
    }
}

############

class Solution {
    public int getMinDistance(int[] nums, int target, int start) {
        int res = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == target) {
                res = Math.min(res, Math.abs(i - start));
            }
        }
        return res;
    }
}

  • // OJ: https://leetcode.com/problems/minimum-distance-to-the-target-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int getMinDistance(vector<int>& A, int target, int start) {
        if (A[start] == target) return 0;
        int i = 1;
        for (int N = A.size(); start + i < N || start - i >= 0; ++i) {
            if ((start + i < N && A[start + i] == target)
               || (start - i >= 0 && A[start - i] == target)) break;
        }
        return i;
    }
};

  • class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        res = inf
        for i, num in enumerate(nums):
            if num == target:
                res = min(res, abs(i - start))
        return res

############

# 1848. Minimum Distance to the Target Element
# https://leetcode.com/problems/minimum-distance-to-the-target-element/

class Solution:
    def getMinDistance(self, nums: List[int], target: int, start: int) -> int:
        res = float('inf')
        
        for i,x in enumerate(nums):
            if x == target:
                res = min(res, abs(i - start))
        
        return res


  • 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
}

  • 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;
}


  • 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()
    }
}

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