# Posts tagged as “prefix sum”

You are given a 0-indexed integer array nums of size n and a positive integer k.

We call an index i in the range k <= i < n - k good if the following conditions are satisfied:

• The k elements that are just before the index i are in non-increasing order.
• The k elements that are just after the index i are in non-decreasing order.

Return an array of all good indices sorted in increasing order.

Example 1:

Input: nums = [2,1,1,1,3,4,1], k = 2
Output: [2,3]
Explanation: There are two good indices in the array:
- Index 2. The subarray [2,1] is in non-increasing order, and the subarray [1,3] is in non-decreasing order.
- Index 3. The subarray [1,1] is in non-increasing order, and the subarray [3,4] is in non-decreasing order.
Note that the index 4 is not good because [4,1] is not non-decreasing.

Example 2:

Input: nums = [2,1,1,2], k = 2
Output: []
Explanation: There are no good indices in this array.


Constraints:

• n == nums.length
• 3 <= n <= 105
• 1 <= nums[i] <= 106
• 1 <= k <= n / 2

## Solution: Prefix Sum

Let before[i] = length of longest non-increasing subarray ends of nums[i].
Let after[i] = length of longest non-decreasing subarray ends of nums[i].

An index is good if nums[i – 1] >= k and nums[i + k] >= k

Time complexity: O(n + (n – 2*k))
Space complexity: O(n)

## C++

You are given a 0-indexed integer array nums of length n.

nums contains a valid split at index i if the following are true:

• The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements.
• There is at least one element to the right of i. That is, 0 <= i < n - 1.

Return the number of valid splits in nums.

Example 1:

Input: nums = [10,4,-8,7]
Output: 2
Explanation:
There are three ways of splitting nums into two non-empty parts:
- Split nums at index 0. Then, the first part is , and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split.
- Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is , and its sum is 7. Since 6 < 7, i = 2 is not a valid split.
Thus, the number of valid splits in nums is 2.


Example 2:

Input: nums = [2,3,1,0]
Output: 2
Explanation:
There are two valid splits in nums:
- Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split.
- Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is , and its sum is 0. Since 6 >= 0, i = 2 is a valid split.


Constraints:

• 2 <= nums.length <= 105
• -105 <= nums[i] <= 105

## Solution: Prefix/Suffix Sum

Note: sum can be greater than 2^31, use long!

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

## C++

You are given a 0-indexed binary array nums of length nnums can be divided at index i (where 0 <= i <= n) into two arrays (possibly empty) numsleft and numsright:

• numsleft has all the elements of nums between index 0 and i - 1 (inclusive), while numsright has all the elements of nums between index i and n - 1 (inclusive).
• If i == 0numsleft is empty, while numsright has all the elements of nums.
• If i == nnumsleft has all the elements of nums, while numsright is empty.

The division score of an index i is the sum of the number of 0‘s in numsleft and the number of 1‘s in numsright.

Return all distinct indices that have the highest possible division score. You may return the answer in any order.

Example 1:

Input: nums = [0,0,1,0]
Output: [2,4]
Explanation: Division at index
- 0: numsleft is []. numsright is [0,0,1,0]. The score is 0 + 1 = 1.
- 1: numsleft is . numsright is [0,1,0]. The score is 1 + 1 = 2.
- 2: numsleft is [0,0]. numsright is [1,0]. The score is 2 + 1 = 3.
- 3: numsleft is [0,0,1]. numsright is . The score is 2 + 0 = 2.
- 4: numsleft is [0,0,1,0]. numsright is []. The score is 3 + 0 = 3.
Indices 2 and 4 both have the highest possible division score 3.
Note the answer [4,2] would also be accepted.

Example 2:

Input: nums = [0,0,0]
Output: 
Explanation: Division at index
- 0: numsleft is []. numsright is [0,0,0]. The score is 0 + 0 = 0.
- 1: numsleft is . numsright is [0,0]. The score is 1 + 0 = 1.
- 2: numsleft is [0,0]. numsright is . The score is 2 + 0 = 2.
- 3: numsleft is [0,0,0]. numsright is []. The score is 3 + 0 = 3.
Only index 3 has the highest possible division score 3.


Example 3:

Input: nums = [1,1]
Output: 
Explanation: Division at index
- 0: numsleft is []. numsright is [1,1]. The score is 0 + 2 = 2.
- 1: numsleft is . numsright is . The score is 0 + 1 = 1.
- 2: numsleft is [1,1]. numsright is []. The score is 0 + 0 = 0.
Only index 0 has the highest possible division score 2.


Constraints:

• n == nums.length
• 1 <= n <= 105
• nums[i] is either 0 or 1.

## Solution: Precompute + Prefix Sum

Count how many ones in the array, track the prefix sum to compute score for each index in O(1).

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

## C++

You are given a 0-indexed array of n integers arr.

The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i - j|.

Return an array intervals of length n where intervals[i] is the sum of intervals between arr[i] and each element in arr with the same value as arr[i].

Note: |x| is the absolute value of x.

Example 1:

Input: arr = [2,1,3,1,2,3,3]
Output: [4,2,7,2,4,4,5]
Explanation:
- Index 0: Another 2 is found at index 4. |0 - 4| = 4
- Index 1: Another 1 is found at index 3. |1 - 3| = 2
- Index 2: Two more 3s are found at indices 5 and 6. |2 - 5| + |2 - 6| = 7
- Index 3: Another 1 is found at index 1. |3 - 1| = 2
- Index 4: Another 2 is found at index 0. |4 - 0| = 4
- Index 5: Two more 3s are found at indices 2 and 6. |5 - 2| + |5 - 6| = 4
- Index 6: Two more 3s are found at indices 2 and 5. |6 - 2| + |6 - 5| = 5


Example 2:

Input: arr = [10,5,10,10]
Output: [5,0,3,4]
Explanation:
- Index 0: Two more 10s are found at indices 2 and 3. |0 - 2| + |0 - 3| = 5
- Index 1: There is only one 5 in the array, so its sum of intervals to identical elements is 0.
- Index 2: Two more 10s are found at indices 0 and 3. |2 - 0| + |2 - 3| = 3
- Index 3: Two more 10s are found at indices 0 and 2. |3 - 0| + |3 - 2| = 4


Constraints:

• n == arr.length
• 1 <= n <= 105
• 1 <= arr[i] <= 105

## Solution: Math / Hashtable + Prefix Sum

For each arr[i], suppose it occurs in the array of total c times, among which k of them are in front of it and c – k – 1 of them are after it. Then the total sum intervals:
(i – j1) + (i – j2) + … + (i – jk) + (jk+1-i) + (jk+2-i) + … + (jc-i)
<=> k * i – sum(j1~jk) + sum(jk+1~jc) – (c – k – 1) * i

Use a hashtable to store the indies of each unique number in the array and compute the prefix sum for fast range sum query.

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

## C++

k x k magic square is a k x k grid filled with integers such that every row sum, every column sum, and both diagonal sums are all equal. The integers in the magic square do not have to be distinct. Every 1 x 1 grid is trivially a magic square.

Given an m x n integer grid, return the size (i.e., the side length k) of the largest magic square that can be found within this grid.

Example 1:

Input: grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]]
Output: 3
Explanation: The largest magic square has a size of 3.
Every row sum, column sum, and diagonal sum of this magic square is equal to 12.
- Row sums: 5+1+6 = 5+4+3 = 2+7+3 = 12
- Column sums: 5+5+2 = 1+4+7 = 6+3+3 = 12
- Diagonal sums: 5+4+3 = 6+4+2 = 12


Example 2:

Input: grid = [[5,1,3,1],[9,3,3,1],[1,3,3,8]]
Output: 2


Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• 1 <= grid[i][j] <= 106

## Solution: Brute Force w/ Prefix Sum

Compute the prefix sum for each row and each column.

And check all possible squares.

Time complexity: O(m*n*min(m,n)2)
Space complexity: O(m*n)