Today I have done Subarray Sums Divisible by K and leetcode's June LeetCoding Challenge with cpp.

Subarray Sums Divisible by K

Description

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

Example 1:

Input: nums = [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], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]

Note:

1. 1 <= nums.length <= 30000
2. -10000 <= nums[i] <= 10000
3. 2 <= k <= 10000

Solution

class Solution {
public:
  int subarraysDivByK(vector<int>& nums, int k) {
    int cnt[k];
    memset(cnt, 0, sizeof(cnt));
    int sum = 0;
    int answer = 0;
    cnt[0] = 1;
    for(auto i : nums) {
      sum = ((sum + i) % k + k) % k;
      answer += cnt[sum];
      cnt[sum] += 1;
    }
    return answer;
  }
};

June LeetCoding Challenge 8

Description

Construct Binary Tree from Preorder and Inorder Traversal

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

Solution

class Solution {
  TreeNode *build(
    vector<int>& preorder,
    int preBegin,
    int preEnd,
    vector<int>& inorder,
    int inBegin,
    int inEnd
  ) {
    if(preBegin > preEnd) return nullptr;
    int rootIdx = -1;
    for(int i = inBegin; i <= inEnd; ++i) {
      if(inorder[i] == preorder[preBegin]) {
        rootIdx = i;
        break;
      }
    }
    int offset = rootIdx - inBegin;
    TreeNode *root = new TreeNode(preorder[preBegin]);
    root->left = build(preorder, preBegin + 1, preBegin + offset, inorder, inBegin, inBegin + offset - 1);
    root->right = build(preorder, preBegin + offset + 1, preEnd, inorder, inBegin + offset + 1, inEnd);
    return root;
  }
public:
  TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
    int len = preorder.size();
    return build(preorder, 0, len - 1, inorder, 0, len - 1);
  }
};