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sureshmangssureshmangs
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🚀 27-Jun-2020
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‎array/array_pair_sum.cpp‎

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/*
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Given an array A of N positive integers and another number X.
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Determine whether or not there exist two elements in A whose sum is exactly X.
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*/
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#include<bits/stdc++.h>
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using namespace std;
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int main()

competitive programming/leetcode/2020-June-Challenge/121. Best Time to Buy and Sell Stock.cpp renamed to competitive programming/leetcode/121. Best Time to Buy and Sell Stock.cpp

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competitive programming/leetcode/2020-June-Challenge/122. Best Time to Buy and Sell Stock II.cpp renamed to competitive programming/leetcode/122. Best Time to Buy and Sell Stock II.cpp

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‎competitive programming/leetcode/129. Sum Root to Leaf Numbers.cpp‎

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Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
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An example is the root-to-leaf path 1->2->3 which represents the number 123.
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Find the total sum of all root-to-leaf numbers.
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Note: A leaf is a node with no children.
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Example:
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Input: [1,2,3]
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1
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/ \
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2 3
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Output: 25
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Explanation:
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The root-to-leaf path 1->2 represents the number 12.
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The root-to-leaf path 1->3 represents the number 13.
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Therefore, sum = 12 + 13 = 25.
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Example 2:
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Input: [4,9,0,5,1]
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4
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/ \
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9 0
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/ \
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5 1
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Output: 1026
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Explanation:
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The root-to-leaf path 4->9->5 represents the number 495.
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The root-to-leaf path 4->9->1 represents the number 491.
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The root-to-leaf path 4->0 represents the number 40.
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Therefore, sum = 495 + 491 + 40 = 1026.
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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long long ans=0;
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void sumNumbersUtil(TreeNode* root, int value){
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if(!root) return;
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value*=10;
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value+=root->val;
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if(!root->left && !root->right){
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ans+=value;
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return;
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}
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sumNumbersUtil(root->left, value);
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sumNumbersUtil(root->right, value);
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}
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int sumNumbers(TreeNode* root) {
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if(!root) return 0;
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sumNumbersUtil(root, 0);
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return ans;
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}
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};

competitive programming/leetcode/2020-June-Challenge/168. Excel Sheet Column Title.cpp renamed to competitive programming/leetcode/168. Excel Sheet Column Title.cpp

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‎competitive programming/leetcode/172. Factorial Trailing Zeroes.cpp‎

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Given an integer n, return the number of trailing zeroes in n!.
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Example 1:
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Input: 3
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Output: 0
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Explanation: 3! = 6, no trailing zero.
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Example 2:
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Input: 5
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Output: 1
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Explanation: 5! = 120, one trailing zero.
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Note: Your solution should be in logarithmic time complexity.
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class Solution {
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public:
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int trailingZeroes(int n) {
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long cnt=0, tmp=5;
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while(tmp<=n){
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cnt+=n/tmp;
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tmp*=5;
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}
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return cnt;
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}
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};

‎competitive programming/leetcode/2020-June-Challenge/Day-26-Sum Root to Leaf Numbers.cpp‎

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Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
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An example is the root-to-leaf path 1->2->3 which represents the number 123.
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Find the total sum of all root-to-leaf numbers.
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Note: A leaf is a node with no children.
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Example:
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Input: [1,2,3]
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1
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/ \
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2 3
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Output: 25
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Explanation:
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The root-to-leaf path 1->2 represents the number 12.
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The root-to-leaf path 1->3 represents the number 13.
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Therefore, sum = 12 + 13 = 25.
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Example 2:
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Input: [4,9,0,5,1]
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4
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/ \
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9 0
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/ \
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5 1
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Output: 1026
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Explanation:
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The root-to-leaf path 4->9->5 represents the number 495.
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The root-to-leaf path 4->9->1 represents the number 491.
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The root-to-leaf path 4->0 represents the number 40.
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Therefore, sum = 495 + 491 + 40 = 1026.
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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long long ans=0;
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void sumNumbersUtil(TreeNode* root, int value){
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if(!root) return;
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value*=10;
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value+=root->val;
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if(!root->left && !root->right){
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ans+=value;
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return;
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}
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sumNumbersUtil(root->left, value);
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sumNumbersUtil(root->right, value);
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}
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int sumNumbers(TreeNode* root) {
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if(!root) return 0;
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sumNumbersUtil(root, 0);
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return ans;
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}
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};

competitive programming/leetcode/2020-June-Challenge/287. Find the Duplicate Number.cpp renamed to competitive programming/leetcode/287. Find the Duplicate Number.cpp

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competitive programming/leetcode/2020-June-Challenge/53. Maximum Subarray.cpp renamed to competitive programming/leetcode/53. Maximum Subarray.cpp

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‎competitive programming/leetcode/62. Unique Paths.cpp‎

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A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
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The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
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How many possible unique paths are there?
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Above is a 7 x 3 grid. How many possible unique paths are there?
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Example 1:
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Input: m = 3, n = 2
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Output: 3
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Explanation:
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From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
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1. Right -> Right -> Down
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2. Right -> Down -> Right
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3. Down -> Right -> Right
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Example 2:
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Input: m = 7, n = 3
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Output: 28
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Constraints:
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1 <= m, n <= 100
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It's guaranteed that the answer will be less than or equal to 2 * 10 ^ 9.
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class Solution {
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public:
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int uniquePaths(int m, int n) {
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int dp[m][n];
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for(int i=0;i<m;i++) dp[i][0]=1;
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for(int j=0;j<n;j++) dp[0][j]=1;
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for(int i=1;i<m;i++){
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for(int j=1;j<n;j++){
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dp[i][j]=dp[i-1][j]+dp[i][j-1];
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}
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}
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return dp[m-1][n-1];
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}
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};

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