Tag: leetcode

class Solution {
public:
    int waysToSplitArray(vector<int>& nums) {
        // Keep track of sum of elements on left and right sides
        long long leftSum = 0, rightSum = 0;

        // Initially all elements are on right side
        for (int num : nums) {
            rightSum += num;
        }

        int count = 0;
        // Try each possible split position
        for (int i = 0; i < nums.size() - 1; i++) {
            // Move current element from right to left side
            leftSum += nums[i];
            rightSum -= nums[i];

            // Check if this creates a valid split
            if (leftSum >= rightSum) {
                count++;
            }
        }

        return count;
    }
};

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 [10], 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 [7], 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 [0], and its sum is 0. Since 6 >= 0, i = 2 is a valid split.

Constraints:

• 2 <= nums.length <= 10<sup>5</sup>
• -10<sup>5</sup> <= nums[i] <= 10<sup>5</sup>

Given the integers zero, one, low, and high, we can construct a string by starting with an empty string, and then at each step perform either of the following:

• Append the character '0' zero times.
• Append the character '1' one times.

This can be performed any number of times.

A good string is a string constructed by the above process having a length between low and high (inclusive).

Return the number of different good strings that can be constructed satisfying these properties. Since the answer can be large, return it modulo 10<sup>9</sup> + 7.

Input: low = 3, high = 3, zero = 1, one = 1
Output: 8
Explanation: 
One possible valid good string is "011". 
It can be constructed as follows: "" -> "0" -> "01" -> "011". 
All binary strings from "000" to "111" are good strings in this example.

Input: low = 2, high = 3, zero = 1, one = 2
Output: 5
Explanation: The good strings are "00", "11", "000", "110", and "011".

Solution: fastest one is using tabulation dynamic programming technique.

    int countGoodStrings(int low, int high, int zero, int one) {
        int sum[100001];
        sum[0] = 1;
        for (int i = 1; i <= high; i++)
        {
            long long sumCur = 0;
            if (i >= zero)
                sumCur += sum[i-zero];
            if (i >= one)
                sumCur += sum[i-one];
            if (sumCur > 0x3000000000000000ll)
                sumCur %= 1000000007;
            sum[i] = sumCur % 1000000007;
        }

        long long sumTotal = 0;
        for (int i = low; i <= high; i++)
            sumTotal += sum[i];
        return sumTotal % 1000000007;
    }

Way slower approach is to use a map in your code for memoization, and a recursive DFS method.

class Solution {
public:
    int countGoodStrings(int low, int high, int zero, int one) {
        map<int, int> dp;
        long mod = pow(10,9) +7;    
        function<int(int)> dfs = [&](int length) -> int{
            
            if( length > high ) return 0;

            if(dp.find(length) != dp.end() )
                return dp[length];

            int count = ( length >= low) ? 1 : 0;
            count = (count + dfs(length + zero)) % mod;
            count = (count + dfs(length + one)) % mod;
            
            return dp[length] = count;
        };

        return dfs(0);    
    }
};

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this

Input: s = "PAYPALISHIRING", numRows = 3 Output: "PAHNAPLSIIGYIR"
P   A   H   N
A P L S I I G
Y   I   R

Input: s = "PAYPALISHIRING", numRows = 4 Output: "PINALSIGYAHRPI"
P     I    N
A   L S  I G
Y A   H R
P     I

Solution

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    string convert(string s, int numRows) {
        int n = s.size();
        if (n == 1 || numRows < 2 || n <= numRows) return s;
        
        string ans;
        
        for (int i = 0; i < numRows; i++) {
            int j = i;
            ans.push_back(s[i]); // First character of the row
            int down = 2 * (numRows - 1 - i); // Downward step size
            int up = 2 * i; // Upward step size
            
            if (up == 0 && down == 0) return s; // If no movement, just return the string
            
            while (j < n) {
                j += down;
                if (j < n && down > 0) ans.push_back(s[j]);
                
                j += up;
                if (j < n && up > 0) ans.push_back(s[j]);
            }
        }
        
        return ans;
    }
};

int main() {
    Solution solution;
    string s = "PAYPALISHIRING"; // Example input
    int numRows = 3; // Example row count
    
    string result = solution.convert(s, numRows);
    cout << "Zigzag Conversion: " << result << endl; // Output the result
    return 0;
}

Given the root of a binary tree, return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

The length of a path between two nodes is represented by the number of edges between them.

Example 1:
Input: root = [1,2,3,4,5] Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].


Example 2:
Input: root = [1,2] Output: 1

Constraints:

-100 <= Node.val <= 100

The number of nodes in the tree is in the range [1, 104].

Solution in C++ with a test case:

#include <iostream>
#include <algorithm>
using namespace std;

// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
public:
    int result;
    int dfs(TreeNode* root) {
        if (root == nullptr) return 0;
        int left = dfs(root->left);
        int right = dfs(root->right);
        result = max(result, left + right);
        return max(left, right) + 1;
    }

    int diameterOfBinaryTree(TreeNode* root) {
        result = 0;
        dfs(root);
        return result;
    }
};

// Helper function to create a tree
TreeNode* createTree() {
    /*
        Example tree:
                1
               / \
              2   3
             / \
            4   5
    */
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    return root;
}

int main() {
    // Create a sample binary tree
    TreeNode* root = createTree();

    // Create a Solution object
    Solution solution;

    // Call the diameterOfBinaryTree function
    int diameter = solution.diameterOfBinaryTree(root);

    // Output the result
    cout << "Diameter of the binary tree: " << diameter << endl;

    // Clean up memory (optional but good practice)
    delete root->left->left;
    delete root->left->right;
    delete root->left;
    delete root->right;
    delete root;

    return 0;
}