Posts tagged as “prefix sum”

花花酱 LeetCode 1769. Minimum Number of Operations to Move All Balls to Each Box

By zxi on February 20, 2021

You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the i^th box is empty, and '1' if it contains one ball.

In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes.

Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the i^th box.

Each answer[i] is calculated considering the initial state of the boxes.

Example 1:

Input: boxes = "110"
Output: [1,1,3]
Explanation: The answer for each box is as follows:
1) First box: you will have to move one ball from the second box to the first box in one operation.
2) Second box: you will have to move one ball from the first box to the second box in one operation.
3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball from the second box to the third box in one operation.

Example 2:

Input: boxes = "001011"
Output: [11,8,5,4,3,4]

Constraints:

n == boxes.length
1 <= n <= 2000
boxes[i] is either '0' or '1'

Solution: Prefix Sum + DP

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

C++

// Author: Huahua

class Solution {

public:

vector<int> minOperations(string boxes) {

const int n = boxes.size();

vector<int> ans(n);

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

ans[i] += c;

c += (s += boxes[i] - '0');

}

for (int i = n - 1, c = 0, s = 0; i >= 0; --i) {

ans[i] += c;

c += (s += boxes[i] - '0');

}

return ans;

}

};

花花酱 LeetCode 1749. Maximum Absolute Sum of Any Subarray

By zxi on February 6, 2021

You are given an integer array nums. The absolute sum of a subarray [nums_l, nums_l+1, ..., nums_r-1, nums_r] is abs(nums_l + nums_l+1 + ... + nums_r-1 + nums_r).

Return the maximum absolute sum of any (possibly empty) subarray of nums.

Note that abs(x) is defined as follows:

If x is a negative integer, then abs(x) = -x.
If x is a non-negative integer, then abs(x) = x.

Example 1:

Input: nums = [1,-3,2,3,-4]
Output: 5
Explanation: The subarray [2,3] has absolute sum = abs(2+3) = abs(5) = 5.

Example 2:

Input: nums = [2,-5,1,-4,3,-2]
Output: 8
Explanation: The subarray [-5,1,-4] has absolute sum = abs(-5+1-4) = abs(-8) = 8.

Constraints:

1 <= nums.length <= 10⁵
-10⁴ <= nums[i] <= 10⁴

Solution: Prefix Sum

ans = max{abs(prefix_sum[i] – max(prefix_sum[0:i])), abs(prefix_sum – min(prefix_sum[0:i])}

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

C++

// Author: Huahua

class Solution {

public:

int maxAbsoluteSum(vector<int>& nums) {

int lo = 0;

int hi = 0;

int s = 0;

int ans = 0;

for (int x : nums) {

s += x;

ans = max({ans, abs(s - lo), abs(s - hi)});

hi = max(hi, s);

lo = min(lo, s);

}

return ans;

}

};

花花酱 LeetCode 1703. Minimum Adjacent Swaps for K Consecutive Ones

By zxi on December 26, 2020

You are given an integer array, nums, and an integer k. nums comprises of only 0‘s and 1‘s. In one move, you can choose two adjacent indices and swap their values.

Return the minimum number of moves required so that nums has k consecutive 1‘s.

Example 1:

Input: nums = [1,0,0,1,0,1], k = 2
Output: 1
Explanation: In 1 move, nums could be [1,0,0,0,1,1] and have 2 consecutive 1's.

Example 2:

Input: nums = [1,0,0,0,0,0,1,1], k = 3
Output: 5
Explanation: In 5 moves, the leftmost 1 can be shifted right until nums = [0,0,0,0,0,1,1,1].

Example 3:

Input: nums = [1,1,0,1], k = 2
Output: 0
Explanation: nums already has 2 consecutive 1's.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is 0 or 1.
1 <= k <= sum(nums)

Solution: Prefix Sum + Sliding Window

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

We only care positions of 1s, we can move one element from position x to y (assuming x + 1 ~ y are all zeros) in y – x steps. e.g. [0 0 1 0 0 0 1] => [0 0 0 0 0 1 1], move first 1 at position 2 to position 5, cost is 5 – 2 = 3.

Given a size k window of indices of ones, the optimal solution it to use the median number as center. We can compute the cost to form consecutive numbers:

e.g. [1 4 7 9 10] => [5 6 7 8 9] cost = (5 – 1) + (6 – 4) + (9 – 8) + (10 – 9) = 8

However, naive solution takes O(n*k) => TLE.

We can use prefix sum to compute the cost of a window in O(1) to reduce time complexity to O(n)

First, in order to use sliding window, we change the target of every number in the window to the median number.
e.g. [1 4 7 9 10] => [7 7 7 7 7] cost = (7 – 1) + (7 – 4) + (7 – 7) + (9 – 7) + (10 – 7) = (9 + 10) – (1 + 4) = right – left.
[5 6 7 8 9] => [7 7 7 7 7] takes extra 2 + 1 + 1 + 2 = 6 steps = (k / 2) * ((k + 1) / 2), these extra steps should be deducted from the final answer.

C++

class Solution {

public:

int minMoves(vector<int>& nums, int k) {

vector<long> s(1);

for (int i = 0; i < nums.size(); ++i)

if (nums[i])

s.push_back(s.back() + i);

const int n = s.size();

long ans = 1e10;

const int m1 = k / 2, m2 = (k + 1) / 2;

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

const long right = s[i + k] - s[i + m1];

const long left = s[i + m2] - s[i];

ans = min(ans, right - left);

}

return ans - m1 * m2;

}

};

Python3

class Solution:

def minMoves(self, nums: List[int], k: int) -> int:

ans = 1e10

s = [0]

for i, v in enumerate(nums):

if v: s.append(s[-1] + i)

n = len(s)

m1 = k // 2

m2 = (k + 1) // 2

for i in range(n - k):

right = s[i + k] - s[i + m1]

left = s[i + m2] - s[i]

ans = min(ans, right - left)

return ans - m1 * m2

花花酱 LeetCode 1685. Sum of Absolute Differences in a Sorted Array

By zxi on December 12, 2020

You are given an integer array nums sorted in non-decreasing order.

Build and return an integer array result with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.

In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).

Example 1:

Input: nums = [2,3,5]
Output: [4,3,5]
Explanation: Assuming the arrays are 0-indexed, then
result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.

Example 2:

Input: nums = [1,4,6,8,10]
Output: [24,15,13,15,21]

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= nums[i + 1] <= 10⁴

Solution: Prefix Sum

Let s[i] denote sum(num[i] – num[j]) 0 <= j <= i
s[i] = s[i – 1] + (num[i] – num[i – 1]) * i
Let l[i] denote sum(nums[j] – nums[i]) i <= j < n
l[i] = l[i + 1] + (nums[i + 1] – num[i]) * (n – i – 1)
ans[i] = s[i] + l[i]

e.g. 1, 3, 7, 9
s[0] = 0
s[1] = 0 + (3 – 1) * 1 = 2
s[2] = 2 + (7 – 3) * 2 = 10
s[3] = 10 + (9 – 7) * 3 = 16
l[3] = 0
l[2] = 0 + (9 – 7) * 1 = 2
l[1] = 2 + (7 – 3) * 2 = 10
l[0] = 10 + (3 – 1) * 3 = 16

ans = [16, 12, 12, 16]

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

C++

// Author: Huahua

class Solution {

public:

vector<int> getSumAbsoluteDifferences(vector<int>& nums) {

const int n = nums.size();

vector<int> ans(n);

for (int i = 1, sum = 0; i < n; ++i)

ans[i] += (sum += (nums[i] - nums[i - 1]) * i);

for (int i = n - 2, sum = 0; i >= 0; --i)

ans[i] += (sum += (nums[i + 1] - nums[i]) * (n - i - 1));

return ans;

}

};

花花酱 LeetCode 1664. Ways to Make a Fair Array

By zxi on November 22, 2020

You are given an integer array nums. You can choose exactly one index (0-indexed) and remove the element. Notice that the index of the elements may change after the removal.

For example, if nums = [6,1,7,4,1]:

Choosing to remove index 1 results in nums = [6,7,4,1].
Choosing to remove index 2 results in nums = [6,1,4,1].
Choosing to remove index 4 results in nums = [6,1,7,4].

An array is fair if the sum of the odd-indexed values equals the sum of the even-indexed values.

Return the number of indices that you could choose such that after the removal, numsis fair.

Example 1:

Input: nums = [2,1,6,4]
Output: 1
Explanation:
Remove index 0: [1,6,4] -> Even sum: 1 + 4 = 5. Odd sum: 6. Not fair.
Remove index 1: [2,6,4] -> Even sum: 2 + 4 = 6. Odd sum: 6. Fair.
Remove index 2: [2,1,4] -> Even sum: 2 + 4 = 6. Odd sum: 1. Not fair.
Remove index 3: [2,1,6] -> Even sum: 2 + 6 = 8. Odd sum: 1. Not fair.
There is 1 index that you can remove to make nums fair.

Example 2:

Input: nums = [1,1,1]
Output: 3
Explanation: You can remove any index and the remaining array is fair.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: You cannot make a fair array after removing any index.

Constraints:

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

Solution: Prefix Sum

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

C++

// Author: Huahua

class Solution {

public:

int waysToMakeFair(vector<int>& nums) {

int n = nums.size();

vector<int> odds(n + 1);

vector<int> evens(n + 1);

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

odds[i + 1] = odds[i] + (i % 2 == 1) * nums[i];

evens[i + 1] = evens[i] + (i % 2 == 0) * nums[i];

}

int ans = 0;

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

const int odd = odds[i] + (evens[n] - evens[i + 1]);

const int even = evens[i] + (odds[n] - odds[i + 1]);

ans += odd == even;

}

return ans;

}

};