Hop Cost - Google Top Interview Questions


Problem Statement :


You are given two lists of integers nums0 and nums1 of the same length as well as integers dist and cost.

You must start off at index 0 at either nums0 or nums1 and want to end up at the last index of either list. 

In each round, you can choose to switch to the other list for cost of cost. 

And then you can jump forward at most dist distance away where the cost of landing at an index is the value at that index.

Return the minimum total cost possible to finish the task.

Constraints

0 ≤ n ≤ 1,000 where n is the length of nums0 and nums1.

dist ≤ 100

Example 1

Input

nums0 = [1, 2, 9, 9, 5]

nums1 = [9, 9, 3, 4, 100]

dist = 2


cost = 3

Output

15

Explanation

We can start off at 1, switch to the second list to 3, and switch back to first list to 5.


The total cost is 1 + 3 + 5 plus the two jumps across lists 3 + 3.



Solution :



title-img




                        Solution in C++ :

const int NTREE = 1024;

void update(vector<int>& tree, int pos, int val) {
    pos += NTREE;
    tree[pos] = val;
    pos /= 2;
    while (pos) {
        tree[pos] = min(tree[2 * pos], tree[2 * pos + 1]);
        pos /= 2;
    }
}

int query(vector<int>& tree, int l, int r) {
    int result = INT_MAX;
    l += NTREE;
    r += NTREE;

    while (l <= r) {
        if (l % 2 == 1) result = min(result, tree[l++]);
        if (r % 2 == 0) result = min(result, tree[r--]);

        l /= 2;
        r /= 2;
    }

    return result;
}

int solve(vector<int>& nums0, vector<int>& nums1, int dist, int cost) {
    int N = nums0.size();
    vector<int> tree0, tree1;
    tree0.resize(2 * NTREE);
    tree1.resize(2 * NTREE);

    for (int i = 0; i < 2 * NTREE; i++) tree0[i] = tree1[i] = INT_MAX;

    update(tree0, 0, nums0[0]);
    update(tree1, 0, nums1[0]);

    for (int i = 1; i < N; i++) {
        int min0 = query(tree0, i - dist, i - 1);
        int min1 = query(tree1, i - dist, i - 1);

        int dp0 = min(min0, min1 + cost) + nums0[i];
        int dp1 = min(min1, min0 + cost) + nums1[i];

        update(tree0, i, dp0);
        update(tree1, i, dp1);
    }

    return min(query(tree0, N - 1, N - 1), query(tree1, N - 1, N - 1));
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums0, int[] nums1, int dist, int cost) {
        int n = nums0.length, a0[] = nums0.clone(), a1[] = nums1.clone();
        for (int i = 1; i < n; i++) {
            a0[i] += a0[i - 1];
            a1[i] += a1[i - 1];
            for (int j = i - 1; j >= 0 && j >= i - dist; j--) {
                a1[i] = Math.min(a1[i], Math.min(a1[j], a0[j] + cost) + nums1[i]);
                a0[i] = Math.min(a0[i], Math.min(a0[j], a1[j] + cost) + nums0[i]);
            }
        }
        return Math.min(a0[n - 1], a1[n - 1]);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums0, nums1, dist, cost):
        n = len(nums0)
        if n == 0:
            return 0
        dp0, dp1 = [0] * n, [0] * n
        monoq0, monoq1 = deque(), deque()
        dp0[0] = nums0[0]
        dp1[0] = nums1[0]
        monoq0.append(nums0[0])
        monoq1.append(nums1[0])
        for i in range(1, n):
            # calculate best costs at current index
            dp0[i] = nums0[i] + min(monoq0[0], monoq1[0] + cost)
            dp1[i] = nums1[i] + min(monoq0[0] + cost, monoq1[0])

            # update monoqueue for dp0
            while len(monoq0) > 0 and monoq0[-1] > dp0[i]:
                monoq0.pop()
            monoq0.append(dp0[i])
            if i >= dist and monoq0[0] == dp0[i - dist]:
                monoq0.popleft()

            # update monoqueue for dp1
            while len(monoq1) > 0 and monoq1[-1] > dp1[i]:
                monoq1.pop()
            monoq1.append(dp1[i])
            if i >= dist and monoq1[0] == dp1[i - dist]:
                monoq1.popleft()

        return min(dp0[n - 1], dp1[n - 1])
                    


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