Day 43.2.txt

1848. Minimum Distance to the Target Element

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.



class Solution {
    public int getMinDistance(int[] n, int t, int s) {
        int i,j,c=Integer.MAX_VALUE,k=0;
        for(i=s;i<n.length;i++)
        {
            if(n[i]==t)
            {
                k=Math.abs(i-s);
                c=Math.min(c,k);
            }
        }
        {
            for(i=s;i>=0;i--)
            {
               if(n[i]==t)
               {
                   k=Math.abs(i-s);
                   c=Math.min(c,k);
               }
            }
        }
        return c;
    }
}