Turn Into Non-Increasing List - Google Top Interview Questions


Problem Statement :


You are given a list of integers nums. 

Consider an operation where we take two consecutive integers and merge it into one by taking their sum. 

Return the minimum number of operations required so that the list becomes non-increasing.

Constraints

n ≤ 1,000 where n is the length of nums.

Example 1

Input

nums = [1, 5, 3, 9, 1]

Output

2

Explanation

We can merge [1, 5] to get [6, 3, 9, 1] and then merge [6, 3] to get [9, 9, 1].



Example 2

Input

nums = [3]

Output

0

Example 3

Input

nums = [9, 9]


Output

0



Solution :



title-img




                        Solution in C++ :

map<int, int> dp[1005];
int solve(vector<int>& v) {
    reverse(v.begin(), v.end());
    int n = v.size();
    for (int i = 0; i <= n; i++) dp[i].clear();
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
        int currsum = 0;
        vector<pair<int, int>> cands;
        for (int j = i; j >= 0; j--) {
            currsum += v[j];
            auto it = dp[j].upper_bound(currsum);
            if (it == dp[j].begin()) continue;
            it--;
            cands.emplace_back(currsum, it->second + 1);
        }
        sort(cands.begin(), cands.end());
        int lowestcount = -1;
        for (auto cand : cands) {
            if (cand.second <= lowestcount) continue;
            dp[i + 1][cand.first] = cand.second;
            lowestcount = cand.second;
        }
    }
    int ret = 0;
    for (auto out : dp[n]) ret = max(ret, out.second);
    return n - ret;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):

        inf = 10 ** 9

        @lru_cache(None)
        def rec(i, maxi):

            if i == len(nums):
                return 0

            sm = 0
            ans = -inf
            for ptr in range(i, len(nums)):
                sm += nums[ptr]
                if sm > maxi:
                    break
                ans = max(ans, rec(ptr + 1, sm) + 1)
                if ans > 0:
                    break

            return ans

        return len(nums) - rec(0, 1000000000)
                    


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