# Posts tagged as “prefix sum”

Given a string s, we make queries on substrings of s.

For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return an array answer[], where answer[i] is the result of the i-th query queries[i].

Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters.  (Also, note that the initial string s is never modified by any query.)

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries : substring = "d", is palidrome.
queries : substring = "bc", is not palidrome.
queries : substring = "abcd", is not palidrome after replacing only 1 character.
queries : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries : substring = "abcda", could be changed to "abcba" which is palidrome.


Constraints:

• 1 <= s.length, queries.length <= 10^5
• 0 <= queries[i] <= queries[i] < s.length
• 0 <= queries[i] <= s.length
• s only contains lowercase English letters.

## Solution: Prefix frequency

Compute the prefix frequency of each characters, then we can efficiently compute the frequency of each characters in the substring in O(1) time. Count the number odd frequency characters o, we can convert it to a palindrome if o / 2 <= k.

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

## C++

In an array A of 0s and 1s, how many non-empty subarrays have sum S?

Example 1:

Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation:
The 4 subarrays are bolded below:
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]


Note:

1. A.length <= 30000
2. 0 <= S <= A.length
3. A[i] is either 0 or 1.

## Solution: Prefix Sum

counts[s] := # of subarrays start from 0 that have sum of s
ans += counts[s – S] if s >= S

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

## C++

Given an array A of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by K.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5 Output: 7 Explanation: There are 7 subarrays with a sum divisible by K = 5: [4, 5, 0, -2, -3, 1], , [5, 0], [5, 0, -2, -3], , [0, -2, -3], [-2, -3]

Note:

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

Solution: Count prefix sums

let c[i] denotes the counts of prefix_sum % K init: c = 1
Whenever we end up with the same prefix sum (after modulo), which means there are subarrys end with current element that is divisible by K (0 modulo).

e.g. A = [4,5,0,-2,-3,1], K = 5
[4,5] has prefix sum of 4, which happens at index 0 , and index 1, [4,5]
[4,5,0] also has a prefix sum of 4, which means [4, {5,0}], [4,5, {0}] are divisible by K.

ans += (c[prefix_sum] – 1)
i = 0, prefix_sum = 0, c[(0+4)%5] = c = 1, ans = 0
i = 1, prefix_sum = 4+5, c[(4+5)%5] = c = 2, ans = 0+2-1=0 => 
i = 2, prefix_sum = 4+0, c[(4+0)%5] = c = 3, ans = 1+3-1=3 => , [5,0], 
i = 3, prefix_sum = 4-2, c[(4-2)%5] = c = 1, ans = 3
i = 4, prefix_sum = 2-3, c[(2-3+5)%5] = c = 4, ans = 3+4-1=6 => ,[5,0],,[5,0,-2,-3], [0,-2,-3],[-2,-3]
i = 5, prefix_sum = 4+1, c[(4+1)%5] = c = 2, ans = 6 + 2 – 1 =>
,[5,0],,[5,0,-2,-3], [0,-2,-3],[-2,-3], [4,5,0,-2,-3,1]

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

# Problem

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3


Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

# Solution: Prefix sum

sums[i] = nums + nums + … + nums[i]

sumRange(i, j) = sums[j] – sums[i – 1]

Time complexity: pre-compute: O(n), query: O(1)

Space complexity: O(n)

# Problem

Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.


Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.


Note:

1. The length of the array won’t exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

# Special case:

nums = [0,0], k = 0, return = True

# Solution: Prefix Sum Reminder

Time complexity: O(n)

Space complexity: O(min(n, k))

# Related Problems

Mission News Theme by Compete Themes.