Crush Consecutive Numbers - Amazon Top Interview Questions

Problem Statement :

You are given a list of positive integers nums and a positive integer k. Consider an operation where you merge exactly k consecutive numbers, which costs the sum of those numbers.

Given that you can make any number of operations, return the minimum cost necessary to merge all the numbers into one number. If it's not possible, return -1.


n ≤ 40 where n is the length of nums

2 ≤ k ≤ 40

nums[i] ≤ 100

Example 1


nums = [1, 2, 4, 3, 2]

k = 3




First we merge [1, 2, 4] into 7 to get [7, 3, 2] for a cost of 7.

Then we merge [7, 3, 2] to get [12] for a cost of 12.

Solution :


                        Solution in C++ :

int solve(vector<int>& nums, int k) {
    int n = nums.size();
    if ((n - 1) % (k - 1) != 0) return -1;

    vector<int> suff(n + 1);
    for (int i = n - 1; i >= 0; i--) {
        suff[i] = nums[i] + suff[i + 1];

    vector<vector<int>> dp(n, vector<int>(n));
    for (int sz = k; sz <= n; sz++) {
        for (int l = 0, r = sz - 1; r < n; l++, r++) {
            dp[l][r] = INT_MAX;
            // merge [l, i] into one pile, and the rest as much as possible
            for (int i = l; i < r; i += k - 1) {
                dp[l][r] = min(dp[l][r], dp[l][i] + dp[i + 1][r]);

            // after doing the above split, the interval [l, r] will be reduced to a size <= k
            // if it's reduced to k, we reduce it again to 1 by adding the final cost
            if ((r - l) % (k - 1) == 0) {
                dp[l][r] += suff[l] - suff[r + 1];

    return dp[0][n - 1];

                        Solution in Python : 
class Solution:
    def solve(self, nums, k):
        dp = [[None] * len(nums) for _ in range(len(nums))]

        def find(i, j):
            if i == j:
                return 0
            if k != 2 and (j - i + 1) % (k - 1) != 1:
                return math.inf
            if dp[i][j] is None:
                ans = sum(nums[i : j + 1])
                min_cost = math.inf
                for split in combinations(range(i + 1, j + 1), k - 1):
                    split = list(split) + [j + 1]
                    its_cost = find(i, split[0] - 1)
                    for ind in range(len(split) - 1):
                        its_cost += find(split[ind], split[ind + 1] - 1)
                    min_cost = min(min_cost, its_cost)
                dp[i][j] = ans + min_cost
            return dp[i][j]

        return find(0, len(nums) - 1) if find(0, len(nums) - 1) != math.inf else -1

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