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3919. Minimum Cost to Move Between Indices

Description

You are given an integer array nums where nums is strictly increasing.

For each index x, let closest(x) be the adjacent index y such that abs(nums[x] - nums[y]) is minimized. If both adjacent indices exist and give the same difference, choose the smaller index.

From any index x, you can move in two ways:

  • To any index y with cost abs(nums[x] - nums[y]), or
  • To closest(x) with cost 1.

You are also given a 2D integer array queries, where each queries[i] = [li, ri].

For each query, calculate the minimum total cost to move from index li to index ri.

Return an integer array ans, where ans[i] is the answer for the ith query.

The absolute difference between two values x and y is defined as abs(x - y).

 

Example 1:

Input: nums = [-5,-2,3], queries = [[0,2],[2,0],[1,2]]

Output: [6,2,5]

Explanation:​​​​​​​​​​​​​​​​​​​​

  • The closest indices are [1, 0, 1] respectively.
  • For [0, 2], the path 0 → 1 → 2 uses a closest move from index 0 to 1 with cost 1 and a move from index 1 to 2 with cost \|-2 - 3\| = 5, giving total 1 + 5 = 6.
  • For [2, 0], the path 2 → 1 → 0 uses two closest moves from index 2 to 1 and from index 1 to 0, each with cost 1, giving total 2.
  • For [1, 2], the direct move from index 1 to index 2 has cost \|-2 - 3\| = 5, which is optimal.

Thus, ans = [6, 2, 5].

Example 2:

Input: nums = [0,2,3,9], queries = [[3,0],[1,2],[2,0]]

Output: [4,1,3]

Explanation:

  • The closest indices are [1, 2, 1, 2] respectively.
  • For [3, 0], the path 3 → 2 → 1 → 0 uses closest moves from index 3 to 2 and from 2 to 1, each with cost 1, and a move from 1 to 0 with cost \|2 - 0\| = 2, giving total 1 + 1 + 2 = 4.
  • For [1, 2], the closest move from index 1 to 2 has cost 1.
  • For [2, 0], the path 2 → 1 → 0 uses a closest move from index 2 to 1 with cost 1 and a move from 1 to 0 with cost \|2 - 0\| = 2, giving total 1 + 2 = 3.

Thus, ans = [4, 1, 3].

 

Constraints:

  • 2 <= nums.length <= 105
  • -109 <= nums[i] <= 109
  • nums is strictly increasing
  • 1 <= queries.length <= 105
  • queries[i] = [li, ri]​​​​​​​
  • 0 <= li, ri < nums.length

Solutions

Solution 1

  • class Solution {
    public int[] minCost(int[] nums, int[][] queries) {
        int n = nums.length;
        int[] s1 = new int[n];
        int[] s2 = new int[n];
        for (int i = 1; i < n; i++) {
            int c1 = (i > 1 && nums[i - 1] - nums[i - 2] <= nums[i] - nums[i - 1])
                ? nums[i] - nums[i - 1]
                : 1;
            int c2 = (i < n - 1 && nums[i] - nums[i - 1] > nums[i + 1] - nums[i])
                ? nums[i] - nums[i - 1]
                : 1;
            s1[i] = s1[i - 1] + c1;
            s2[i] = s2[i - 1] + c2;
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; i++) {
            int l = queries[i][0];
            int r = queries[i][1];
            ans[i] = (l < r) ? s1[r] - s1[l] : s2[l] - s2[r];
        }
        return ans;
    }
}


  • class Solution {
public:
    vector<int> minCost(vector<int>& nums, vector<vector<int>>& queries) {
        int n = nums.size();
        vector<int> s1(n, 0);
        vector<int> s2(n, 0);
        for (int i = 1; i < n; i++) {
            int c1 = (i > 1 && nums[i - 1] - nums[i - 2] <= nums[i] - nums[i - 1]) ? nums[i] - nums[i - 1] : 1;
            int c2 = (i < n - 1 && nums[i] - nums[i - 1] > nums[i + 1] - nums[i]) ? nums[i] - nums[i - 1] : 1;
            s1[i] = s1[i - 1] + c1;
            s2[i] = s2[i - 1] + c2;
        }
        int m = queries.size();
        vector<int> ans(m);
        for (int i = 0; i < m; i++) {
            int l = queries[i][0];
            int r = queries[i][1];
            ans[i] = (l < r) ? s1[r] - s1[l] : s2[l] - s2[r];
        }
        return ans;
    }
};


  • class Solution:
    def minCost(self, nums: list[int], queries: list[list[int]]) -> list[int]:
        n = len(nums)
        s1 = [0] * n
        s2 = [0] * n
        for i in range(1, n):
            c1 = (
                nums[i] - nums[i - 1]
                if i > 1 and nums[i - 1] - nums[i - 2] <= nums[i] - nums[i - 1]
                else 1
            )
            c2 = (
                nums[i] - nums[i - 1]
                if i < n - 1 and nums[i] - nums[i - 1] > nums[i + 1] - nums[i]
                else 1
            )
            s1[i] = s1[i - 1] + c1
            s2[i] = s2[i - 1] + c2
        m = len(queries)
        ans = [0] * m
        for i, (l, r) in enumerate(queries):
            ans[i] = s1[r] - s1[l] if l < r else s2[l] - s2[r]
        return ans


  • func minCost(nums []int, queries [][]int) []int {
	n := len(nums)
	s1 := make([]int, n)
	s2 := make([]int, n)
	for i := 1; i < n; i++ {
		c1 := 1
		if i > 1 && nums[i-1]-nums[i-2] <= nums[i]-nums[i-1] {
			c1 = nums[i] - nums[i-1]
		}
		c2 := 1
		if i < n-1 && nums[i]-nums[i-1] > nums[i+1]-nums[i] {
			c2 = nums[i] - nums[i-1]
		}
		s1[i] = s1[i-1] + c1
		s2[i] = s2[i-1] + c2
	}
	m := len(queries)
	ans := make([]int, m)
	for i := 0; i < m; i++ {
		l := queries[i][0]
		r := queries[i][1]
		if l < r {
			ans[i] = s1[r] - s1[l]
		} else {
			ans[i] = s2[l] - s2[r]
		}
	}
	return ans
}


  • function minCost(nums: number[], queries: number[][]): number[] {
    const n = nums.length;
    const s1: number[] = new Array(n).fill(0);
    const s2: number[] = new Array(n).fill(0);
    for (let i = 1; i < n; i++) {
        const c1 =
            i > 1 && nums[i - 1] - nums[i - 2] <= nums[i] - nums[i - 1] ? nums[i] - nums[i - 1] : 1;
        const c2 =
            i < n - 1 && nums[i] - nums[i - 1] > nums[i + 1] - nums[i] ? nums[i] - nums[i - 1] : 1;
        s1[i] = s1[i - 1] + c1;
        s2[i] = s2[i - 1] + c2;
    }
    const m = queries.length;
    const ans: number[] = new Array(m);
    for (let i = 0; i < m; i++) {
        const l = queries[i][0];
        const r = queries[i][1];
        ans[i] = l < r ? s1[r] - s1[l] : s2[l] - s2[r];
    }
    return ans;
}

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