Posts tagged as “prefix sum”

花花酱 LeetCode 1658. Minimum Operations to Reduce X to Zero

By zxi on November 14, 2020

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it’s possible, otherwise, return -1.

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁴
1 <= x <= 10⁹

Solution1: Prefix Sum + Hashtable

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int minOperations(vector<int>& nums, int x) {

const int n = nums.size();

vector<int> l(n), r(n);

unordered_map<int, int> ml, mr;

ml[0] = mr[0] = -1;

for (int i = 0; i < n; ++i) {

l[i] = nums[i] + (i > 0 ? l[i - 1] : 0);

r[i] = nums[n - i - 1] + (i > 0 ? r[i - 1]: 0);

ml[l[i]] = mr[r[i]] = i;

}

int ans = INT_MAX;

for (int i = 0; i < n; ++i) {

int s1 = x - l[i];

auto it1 = mr.find(s1);

if (it1 != mr.end()) ans = min(ans, i + 1 + it1->second + 1);

int s2 = x - r[i];

auto it2 = ml.find(s2);

if (it2 != ml.end()) ans = min(ans, i + 1 + it2->second + 1);

}

return ans > n ? -1 : ans;

}

};

Solution2: Sliding Window

Find the longest sliding window whose sum of elements equals sum(nums) – x
ans = n – window_size

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int minOperations(vector<int>& nums, int x) {

const int n = nums.size();

int target = accumulate(begin(nums), end(nums), 0) - x;

int ans = INT_MAX;

for (int s = 0, l = 0, r = 0; r < n; ++r) {

s += nums[r];

while (s > target && l <= r) s -= nums[l++];

if (s == target) ans = min(ans, n - (r - l + 1));

}

return ans > n ? -1 : ans;

}

};

花花酱 LeetCode 1649. Create Sorted Array through Instructions

By zxi on November 8, 2020

Given an integer array instructions, you are asked to create a sorted array from the elements in instructions. You start with an empty container nums. For each element from left to right in instructions, insert it into nums. The cost of each insertion is the minimum of the following:

The number of elements currently in nums that are strictly less than instructions[i].
The number of elements currently in nums that are strictly greater than instructions[i].

For example, if inserting element 3 into nums = [1,2,3,5], the cost of insertion is min(2, 1) (elements 1 and 2 are less than 3, element 5 is greater than 3) and nums will become [1,2,3,3,5].

Return the total cost to insert all elements from instructions into nums. Since the answer may be large, return it modulo 10⁹ + 7

Example 1:

Input: instructions = [1,5,6,2]
Output: 1
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 5 with cost min(1, 0) = 0, now nums = [1,5].
Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].
Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].
The total cost is 0 + 0 + 0 + 1 = 1.

Example 2:

Input: instructions = [1,2,3,6,5,4]
Output: 3
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 2 with cost min(1, 0) = 0, now nums = [1,2].
Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].
Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].
Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].
Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].
The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.

Example 3:

Input: instructions = [1,3,3,3,2,4,2,1,2]
Output: 4
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].
Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].
Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].
Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].
Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].
Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].
The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.

Constraints:

1 <= instructions.length <= 10⁵
1 <= instructions[i] <= 10⁵

Solution: Fenwick Tree / Binary Indexed Tree

Time complexity: O(nlogm)
Space complexity: O(m + n)

m is the maximum value, n is number of values.

C++

class FenwickTree {

public:

FenwickTree(int n): sums_(n + 1, 0) {}

void update(int i, int delta) {

while (i < sums_.size()) {

sums_[i] += delta;

i += lowbit(i);

}

int query(int i) const {

int sum = 0;

while (i > 0) {

sum += sums_[i];

i -= lowbit(i);

}

return sum;

}

private:

static inline int lowbit(int x) {

return x & (-x);

}

vector<int> sums_;

};

class Solution {

public:

int createSortedArray(vector<int>& instructions) {

const int n = instructions.size();

FenwickTree tree(1e5);

long ans = 0;

for (int i = 0; i < n; ++i) {

int lt = tree.query(instructions[i] - 1);

int gt = i - tree.query(instructions[i]);

ans += min(lt, gt);

tree.update(instructions[i], 1);

}

return ans % 1000000007;

}

};

花花酱 LeetCode 1590. Make Sum Divisible by P

By zxi on September 20, 2020

Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.

Return the length of the smallest subarray that you need to remove, or -1 if it’s impossible.

A subarray is defined as a contiguous block of elements in the array.

Example 1:

Input: nums = [3,1,4,2], p = 6
Output: 1
Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

Example 2:

Input: nums = [6,3,5,2], p = 9
Output: 2
Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

Example 3:

Input: nums = [1,2,3], p = 3
Output: 0
Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

Example 4:

Input: nums = [1,2,3], p = 7
Output: -1
Explanation: There is no way to remove a subarray in order to get a sum divisible by 7.

Example 5:

Input: nums = [1000000000,1000000000,1000000000], p = 3
Output: 0

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= p <= 10⁹

Solution: HashTable + Prefix Sum

Very similar to subarray target sum.

Basically, we are trying to find a shortest subarray that has sum % p equals to r = sum(arr) % p.

We use a hashtable to store the last index of the prefix sum % p and check whether (prefix_sum + p – r) % p exists or not.

Time complexity: O(n)
Space complexity: O(n)

C++

class Solution {

public:

int minSubarray(vector<int>& nums, int p) {

const int n = nums.size();

int r = accumulate(begin(nums), end(nums), 0LL) % p;

if (r == 0) return 0;

unordered_map<int, int> m{{0, -1}}; // {prefix_sum % p -> last_index}

int s = 0;

int ans = n;

for (int i = 0; i < n; ++i) {

s = (s + nums[i]) % p;

auto it = m.find((s + p - r) % p);

if (it != m.end())

ans = min(ans, i - it->second);

m[s] = i;

}

return ans == n ? -1 : ans;

}

};

Python3

class Solution:

def minSubarray(self, nums: List[int], p: int) -> int:

r = sum(nums) % p

if r == 0: return 0

m = {0: -1}

ans = len(nums)

s = 0

for i, x in enumerate(nums):

s = (s + x) % p

t = (s + p - r) % p

if t in m:

ans = min(ans, i - m[t])

m[s] = i

return -1 if ans == len(nums) else ans

花花酱 LeetCode 1588. Sum of All Odd Length Subarrays

By zxi on September 19, 2020

Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.

A subarray is a contiguous subsequence of the array.

Return the sum of all odd-length subarrays of arr.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

1 <= arr.length <= 100
1 <= arr[i] <= 1000

Solution 0: Brute Force

Enumerate all odd length subarrys: O(n^2), each take O(n) to compute the sum.

Total time complexity: O(n^3)
Space complexity: O(1)

Solution 1: Running Prefix Sum

Reduce the time complexity to O(n^2)

C++

class Solution {

public:

int sumOddLengthSubarrays(vector<int>& arr) {

const int n = arr.size();

int ans = 0;

for (int i = 0, s = 0; i < n; ++i, s = 0)

for (int j = i; j < n; ++j) {

s += arr[j];

ans += s * ((j - i + 1) & 1);

}

return ans;

}

};

Solution 2: Math

Count how many times arr[i] can be in of an odd length subarray
we chose the start, which can be 0, 1, 2, … i, i + 1 choices
we chose the end, which can be i, i + 1, … n – 1, n – i choices
Among those 1/2 are odd length.
So there will be upper((i + 1) * (n – i) / 2) odd length subarrays contain arr[i]

ans = sum(((i + 1) * (n – i) + 1) / 2 * arr[i] for in range(n))

Time complexity: O(n)
Space complexity: O(1)

C++

class Solution {

public:

int sumOddLengthSubarrays(vector<int>& arr) {

const int n = arr.size();

int ans = 0;

for (int i = 0; i < n; ++i)

ans += ((i + 1) * (n - i) + 1) / 2 * arr[i];

return ans;

}

};

花花酱 LeetCode 1563. Stone Game V

By zxi on August 23, 2020

There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.

In each round of the game, Alice divides the row into two non-empty rows (i.e. left row and right row), then Bob calculates the value of each row which is the sum of the values of all the stones in this row. Bob throws away the row which has the maximum value, and Alice’s score increases by the value of the remaining row. If the value of the two rows are equal, Bob lets Alice decide which row will be thrown away. The next round starts with the remaining row.

The game ends when there is only one stone remaining. Alice’s is initially zero.

Return the maximum score that Alice can obtain.

Example 1:

Input: stoneValue = [6,2,3,4,5,5]
Output: 18
Explanation: In the first round, Alice divides the row to [6,2,3], [4,5,5]. The left row has the value 11 and the right row has value 14. Bob throws away the right row and Alice's score is now 11.
In the second round Alice divides the row to [6], [2,3]. This time Bob throws away the left row and Alice's score becomes 16 (11 + 5).
The last round Alice has only one choice to divide the row which is [2], [3]. Bob throws away the right row and Alice's score is now 18 (16 + 2). The game ends because only one stone is remaining in the row.

Example 2:

Input: stoneValue = [7,7,7,7,7,7,7]
Output: 28

Example 3:

Input: stoneValue = [4]
Output: 0

Constraints:

1 <= stoneValue.length <= 500
1 <= stoneValue[i] <= 10^6

Solution: Range DP + Prefix Sum

dp[l][r] := max store Alice can get from range [l, r]
sum_l = sum(l, k), sum_r = sum(k + 1, r)
dp[l][r] = max{
dp[l][k] + sum_l if sum_l < sum_r
dp[k+1][r] + sum_r if sum_r < sum_l
max(dp[l][k], dp[k+1][r])) + sum_l if sum_l == sum_r)
} for k in [l, r)

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

class Solution {

public:

int stoneGameV(vector<int>& stoneValue) {

const int n = stoneValue.size();

vector<int> sums(n + 1);

for (int i = 0; i < n; ++i)

sums[i + 1] = sums[i] + stoneValue[i];

vector<vector<int>> cache(n, vector<int>(n, -1));

// max value alice can get from range [l, r]

function<int(int, int)> dp = [&](int l, int r) {

if (l == r) return 0;

int& ans = cache[l][r];

if (ans != -1) return ans;

for (int k = l; k < r; ++k) {

// left: [l, k], right: [k + 1, r]

int sum_l = sums[k + 1] - sums[l];

int sum_r = sums[r + 1] - sums[k + 1];

if (sum_l > sum_r)

ans = max(ans, sum_r + dp(k + 1, r));

else if (sum_l < sum_r)

ans = max(ans, sum_l + dp(l, k));

else

ans = max(ans, sum_l + max(dp(l, k), dp(k + 1, r)));

}

return ans;

};

return dp(0, n - 1);

}

};