Posts tagged as “array”

Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs(i, j)where0 <= i < j < nandnums[i] + nums[j] < target.

Example 1:

Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation: There are 3 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
- (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.


Example 2:

Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation: There are 10 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
- (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
- (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
- (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
- (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
- (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
- (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
- (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
- (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target


Constraints:

• 1 <= nums.length == n <= 50
• -50 <= nums[i], target <= 50

Solution 1: Brute force

Enumerate all pairs.

Time complexity: O(n2)
Space complexity: O(1)

Solution 2: Two Pointers

Sort the numbers.

Use two pointers i, and j.
Set i to 0 and j to n – 1.
while (nums[i] + nums[j] >= target) –j
then we have nums[i] + nums[k] < target (i < k <= j), in total (j – i) pairs.
++i, move to the next starting number.
Time complexity: O(nlogn + n)
Space complexity: O(1)

C++

You are given a 0-indexed integer array nums. You have to find the maximum sum of a pair of numbers from nums such that the maximum digit in both numbers are equal.

Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [51,71,17,24,42]
Output: 88
Explanation:
For i = 1 and j = 2, nums[i] and nums[j] have equal maximum digits with a pair sum of 71 + 17 = 88.
For i = 3 and j = 4, nums[i] and nums[j] have equal maximum digits with a pair sum of 24 + 42 = 66.
It can be shown that there are no other pairs with equal maximum digits, so the answer is 88.

Example 2:

Input: nums = [1,2,3,4]
Output: -1
Explanation: No pair exists in nums with equal maximum digits.


Constraints:

• 2 <= nums.length <= 100
• 1 <= nums[i] <= 104

Solution: Brute Force

Enumerate all pairs of nums and check their sum and max digit.

Time complexity: O(n2)
Space complexity: O(1)

C++

You are given two 0-indexed integer arrays nums1 and nums2 of length n.

Let’s define another 0-indexed integer array, nums3, of length n. For each index i in the range [0, n - 1], you can assign either nums1[i] or nums2[i] to nums3[i].

Your task is to maximize the length of the longest non-decreasing subarray in nums3 by choosing its values optimally.

Return an integer representing the length of the longest non-decreasing subarray in nums3.

Note: subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums1 = [2,3,1], nums2 = [1,2,1]
Output: 2
Explanation: One way to construct nums3 is:
nums3 = [nums1[0], nums2[1], nums2[2]] => [2,2,1].
The subarray starting from index 0 and ending at index 1, [2,2], forms a non-decreasing subarray of length 2.
We can show that 2 is the maximum achievable length.

Example 2:

Input: nums1 = [1,3,2,1], nums2 = [2,2,3,4]
Output: 4
Explanation: One way to construct nums3 is:
nums3 = [nums1[0], nums2[1], nums2[2], nums2[3]] => [1,2,3,4].
The entire array forms a non-decreasing subarray of length 4, making it the maximum achievable length.


Example 3:

Input: nums1 = [1,1], nums2 = [2,2]
Output: 2
Explanation: One way to construct nums3 is:
nums3 = [nums1[0], nums1[1]] => [1,1].
The entire array forms a non-decreasing subarray of length 2, making it the maximum achievable length.


Constraints:

• 1 <= nums1.length == nums2.length == n <= 105
• 1 <= nums1[i], nums2[i] <= 109

Solution: DP

Let dp1(i), dp2(i) denote the length of the Longest Non-decreasing Subarray ends with nums1[i] and nums2[i] respectively.

init: dp1(0) = dp2(0) = 1

dp1(i) = max(dp1(i – 1) + 1 if nums1[i] >= nums1[i – 1] else 1, dp2(i – 1) + 1 if nums1[i] >= nums2[i – 1] else 1)
dp2(i) = max(dp1(i – 1) + 1 if nums2[i] >= nums1[i – 1] else 1, dp2(i – 1) + 1 if nums2[i] >= nums2[i – 1] else 1)

ans = max(dp1, dp2)

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

C++

You are given two 0-indexed integer permutations A and B of length n.

prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B.

Return the prefix common array of A and B.

A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Example 1:

Input: A = [1,3,2,4], B = [3,1,2,4]
Output: [0,2,3,4]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: 1 and 3 are common in A and B, so C[1] = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.


Example 2:

Input: A = [2,3,1], B = [3,1,2]
Output: [0,1,3]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: only 3 is common in A and B, so C[1] = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.


Constraints:

• 1 <= A.length == B.length == n <= 50
• 1 <= A[i], B[i] <= n
• It is guaranteed that A and B are both a permutation of n integers.

Solution 1: Bitset

Use bitsets to store the numbers seen so far for each array, and use sA & sB to count the common elements.

Time complexity: O(n*50)
Space complexity: O(50)

Solution 2: Counter

Use a counter to track the frequency of each element, when the counter[x] == 2, we found a pair.

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

C++

We define the conversion array conver of an array arr as follows:

• conver[i] = arr[i] + max(arr[0..i]) where max(arr[0..i]) is the maximum value of arr[j] over 0 <= j <= i.

We also define the score of an array arr as the sum of the values of the conversion array of arr.

Given a 0-indexed integer array nums of length n, return an array ans of length n where ans[i] is the score of the prefix nums[0..i].

Example 1:

Input: nums = [2,3,7,5,10]
Output: [4,10,24,36,56]
Explanation:
For the prefix [2], the conversion array is [4] hence the score is 4
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56


Example 2:

Input: nums = [1,1,2,4,8,16]
Output: [2,4,8,16,32,64]
Explanation:
For the prefix [1], the conversion array is [2] hence the score is 2
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64


Constraints:

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

Solution: Prefix Sum

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