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 :
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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