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🚀 29-Aug-2020
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‎competitive programming/leetcode/1143. Longest Common Subsequence.cpp

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Given two strings text1 and text2, return the length of their longest common subsequence.
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A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
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If there is no common subsequence, return 0.
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Example 1:
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Input: text1 = "abcde", text2 = "ace"
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Output: 3
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Explanation: The longest common subsequence is "ace" and its length is 3.
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Example 2:
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Input: text1 = "abc", text2 = "abc"
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Output: 3
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Explanation: The longest common subsequence is "abc" and its length is 3.
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Example 3:
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Input: text1 = "abc", text2 = "def"
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Output: 0
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Explanation: There is no such common subsequence, so the result is 0.
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Constraints:
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1 <= text1.length <= 1000
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1 <= text2.length <= 1000
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The input strings consist of lowercase English characters only.
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class Solution {
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public:
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int LCS(string text1, string text2, int m, int n){
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int dp[m+1][n+1];
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for(int i=0;i<m+1;i++) dp[i][0]=0;
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for(int j=0;j<n+1;j++) dp[0][j]=0;
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for(int i=1;i<m+1;i++){
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for(int j=1;j<n+1;j++){
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if(text1[i-1]==text2[j-1])
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dp[i][j]=1+dp[i-1][j-1];
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else dp[i][j]=max(dp[i-1][j], dp[i][j-1]);
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}
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}
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return dp[m][n];
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}
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int longestCommonSubsequence(string text1, string text2) {
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int m=text1.length();
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int n=text2.length();
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return LCS(text1, text2, m, n);
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}
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};

‎competitive programming/leetcode/2020-August-Challenge/Day-28-Implement Rand10() Using Rand7().cpp

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Given the API rand7 which generates a uniform random integer in the range 1 to 7, write a function rand10 which generates a uniform random integer in the range 1 to 10. You can only call the API rand7 and you shouldn't call any other API. Please don't use the system's Math.random().
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Notice that Each test case has one argument n, the number of times that your implemented function rand10 will be called while testing.
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Follow up:
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What is the expected value for the number of calls to rand7() function?
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Could you minimize the number of calls to rand7()?
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Example 1:
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Input: n = 1
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Output: [2]
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Example 2:
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Input: n = 2
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Output: [2,8]
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Example 3:
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Input: n = 3
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Output: [3,8,10]
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Constraints:
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1 <= n <= 105
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// The rand7() API is already defined for you.
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// int rand7();
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// @return a random integer in the range 1 to 7
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class Solution {
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public:
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int rand10() {
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while(1){
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int row=(rand7()-1)*7;
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int col=rand7();
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int curr=row+col;
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if(curr<=40)
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return (curr-1)%10 + 1;
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}
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}
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};
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/*
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rand7() gives a random number in range of 1 to 7
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rand7()-1 gives a random number in range of 0 to 6
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(rand7()-1)*7 gives a random number in range of 0 to 42
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For a uniform distribution we want only numbers in range of 0 to 40
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*/

‎competitive programming/leetcode/300. Longest Increasing Subsequence.cpp

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Given an unsorted array of integers, find the length of longest increasing subsequence.
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Example:
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Input: [10,9,2,5,3,7,101,18]
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Output: 4
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Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
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Note:
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There may be more than one LIS combination, it is only necessary for you to return the length.
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Your algorithm should run in O(n2) complexity.
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Follow up: Could you improve it to O(n log n) time complexity?
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// O(n^2)
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class Solution {
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public:
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int lengthOfLIS(vector<int>& nums) {
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int n=nums.size();
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if(n==0) return 0;
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vector<int> lis(n,1);
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for(int i=1;i<n;i++){
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for(int j=0; j<i; j++){
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if(nums[i]>nums[j] && lis[i] < lis[j]+1)
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lis[i]=lis[j]+1;
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}
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}
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return *max_element(lis.begin(), lis.end());
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}
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};
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// O(nlogn)
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class Solution {
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public:
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int lengthOfLIS(vector<int>& nums) {
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int n=nums.size();
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if(n==0) return 0;
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vector<int> seq;
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seq.push_back(nums[0]);
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for(int i=1;i<n;i++){
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if(seq.back()<nums[i])
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seq.push_back(nums[i]);
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else {
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int index=lower_bound(seq.begin(), seq.end(), nums[i])-seq.begin();
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seq[index]=nums[i];
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}
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}
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return seq.size();
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}
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};

‎competitive programming/leetcode/470. Implement Rand10() Using Rand7().cpp

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Given the API rand7 which generates a uniform random integer in the range 1 to 7, write a function rand10 which generates a uniform random integer in the range 1 to 10. You can only call the API rand7 and you shouldn't call any other API. Please don't use the system's Math.random().
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Notice that Each test case has one argument n, the number of times that your implemented function rand10 will be called while testing.
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Follow up:
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What is the expected value for the number of calls to rand7() function?
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Could you minimize the number of calls to rand7()?
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Example 1:
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Input: n = 1
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Output: [2]
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Example 2:
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Input: n = 2
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Output: [2,8]
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Example 3:
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Input: n = 3
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Output: [3,8,10]
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Constraints:
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1 <= n <= 105
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// The rand7() API is already defined for you.
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// int rand7();
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// @return a random integer in the range 1 to 7
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class Solution {
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public:
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int rand10() {
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while(1){
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int row=(rand7()-1)*7;
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int col=rand7();
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int curr=row+col;
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if(curr<=40)
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return (curr-1)%10 + 1;
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}
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}
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};
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/*
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rand7() gives a random number in range of 1 to 7
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rand7()-1 gives a random number in range of 0 to 6
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(rand7()-1)*7 gives a random number in range of 0 to 42
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For a uniform distribution we want only numbers in range of 0 to 40
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*/

‎competitive programming/leetcode/72. Edit Distance.cpp

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Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
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You have the following 3 operations permitted on a word:
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Insert a character
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Delete a character
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Replace a character
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Example 1:
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Input: word1 = "horse", word2 = "ros"
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Output: 3
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Explanation:
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horse -> rorse (replace 'h' with 'r')
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rorse -> rose (remove 'r')
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rose -> ros (remove 'e')
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Example 2:
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Input: word1 = "intention", word2 = "execution"
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Output: 5
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Explanation:
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intention -> inention (remove 't')
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inention -> enention (replace 'i' with 'e')
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enention -> exention (replace 'n' with 'x')
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exention -> exection (replace 'n' with 'c')
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exection -> execution (insert 'u')
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// Recursive (TLE)
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class Solution {
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public:
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int editDistance(string word1, string word2, int m, int n){
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if(m==0) return n;
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if(n==0) return m;
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if(word1[m-1]==word2[n-1]) return editDistance(word1, word2, m-1, n-1);
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return 1 + min(editDistance(word1, word2, m-1, n), min(editDistance(word1, word2, m, n-1), editDistance(word1, word2, m-1, n-1)));
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}
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int minDistance(string word1, string word2) {
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int m=word1.length();
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int n=word2.length();
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return editDistance(word1, word2, m, n);
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}
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};
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// Recursive (memoized)
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class Solution {
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public:
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int editDistance(string word1, string word2, int m, int n, vector<vector<int> > &dp){
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if(m==0) return n;
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if(n==0) return m;
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if(dp[m-1][n-1]!=-1) return dp[m-1][n-1];
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if(word1[m-1]==word2[n-1]) return dp[m-1][n-1]=editDistance(word1, word2, m-1, n-1, dp);
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return dp[m-1][n-1]= 1 + min(editDistance(word1, word2, m-1, n, dp), min(editDistance(word1, word2, m, n-1, dp), editDistance(word1, word2, m-1, n-1, dp)));
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}
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int minDistance(string word1, string word2) {
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int m=word1.length();
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int n=word2.length();
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vector<vector<int> > dp(m+1, vector<int> (n+1, -1));
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return editDistance(word1, word2, m, n, dp);
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}
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};
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// DP
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class Solution {
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public:
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int editDistance(string word1, string word2, int m, int n){
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int dp[m+1][n+1];
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for(int i=0;i<m+1;i++) dp[i][0]=i;
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for(int j=0;j<n+1;j++) dp[0][j]=j;
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for(int i=1;i<m+1;i++){
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for(int j=1;j<n+1;j++){
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if(word1[i-1]==word2[j-1]) dp[i][j]=dp[i-1][j-1];
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else dp[i][j]= 1+ min(dp[i-1][j],min(dp[i][j-1],dp[i-1][j-1]));
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}
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}
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return dp[m][n];
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}
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int minDistance(string word1, string word2) {
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int m=word1.length();
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int n=word2.length();
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return editDistance(word1, word2, m, n);
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}
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};

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