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🚀 14-Aug-2020
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‎competitive programming/leetcode/1286. Iterator for Combination.cpp

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Design an Iterator class, which has:
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A constructor that takes a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments.
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A function next() that returns the next combination of length combinationLength in lexicographical order.
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A function hasNext() that returns True if and only if there exists a next combination.
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Example:
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CombinationIterator iterator = new CombinationIterator("abc", 2); // creates the iterator.
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iterator.next(); // returns "ab"
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iterator.hasNext(); // returns true
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iterator.next(); // returns "ac"
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iterator.hasNext(); // returns true
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iterator.next(); // returns "bc"
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iterator.hasNext(); // returns false
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Constraints:
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1 <= combinationLength <= characters.length <= 15
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There will be at most 10^4 function calls per test.
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It's guaranteed that all calls of the function next are valid.
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class CombinationIterator {
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public:
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queue<string> q;
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void allCombinations(string characters, int k, int s, string str){
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if(k==0){
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q.push(str);
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return;
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}
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for(int i=s; i<characters.length();i++){
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str+=characters[i];
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allCombinations(characters, k-1, i+1, str);
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str.erase(str.end()-1);
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}
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}
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CombinationIterator(string characters, int combinationLength) {
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string str="";
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allCombinations(characters, combinationLength, 0, str); // <characters, combinationLength, start, str>
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}
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string next() {
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string next = q.front();
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q.pop();
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return next;
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}
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bool hasNext() {
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return !q.empty();
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}
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};
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/**
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* Your CombinationIterator object will be instantiated and called as such:
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* CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
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* string param_1 = obj->next();
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* bool param_2 = obj->hasNext();
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*/

‎competitive programming/leetcode/2020-August-Challenge/Day-13-Iterator for Combination.cpp

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Design an Iterator class, which has:
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A constructor that takes a string characters of sorted distinct lowercase English letters and a number combinationLength as arguments.
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A function next() that returns the next combination of length combinationLength in lexicographical order.
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A function hasNext() that returns True if and only if there exists a next combination.
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Example:
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CombinationIterator iterator = new CombinationIterator("abc", 2); // creates the iterator.
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iterator.next(); // returns "ab"
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iterator.hasNext(); // returns true
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iterator.next(); // returns "ac"
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iterator.hasNext(); // returns true
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iterator.next(); // returns "bc"
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iterator.hasNext(); // returns false
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Constraints:
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1 <= combinationLength <= characters.length <= 15
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There will be at most 10^4 function calls per test.
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It's guaranteed that all calls of the function next are valid.
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Hide Hint #1
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Generate all combinations as a preprocessing.
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Hide Hint #2
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Use bit masking to generate all the combinations.
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class CombinationIterator {
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public:
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queue<string> q;
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void allCombinations(string characters, int k, int s, string str){
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if(k==0){
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q.push(str);
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return;
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}
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for(int i=s; i<characters.length();i++){
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str+=characters[i];
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allCombinations(characters, k-1, i+1, str);
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str.erase(str.end()-1);
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}
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}
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CombinationIterator(string characters, int combinationLength) {
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string str="";
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allCombinations(characters, combinationLength, 0, str); // <characters, combinationLength, start, str>
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}
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string next() {
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string next = q.front();
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q.pop();
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return next;
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}
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bool hasNext() {
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return !q.empty();
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}
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};
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/**
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* Your CombinationIterator object will be instantiated and called as such:
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* CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
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* string param_1 = obj->next();
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* bool param_2 = obj->hasNext();
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*/

‎competitive programming/leetcode/2020-August-Challenge/Day-14-Longest Palindrome.cpp

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Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
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This is case sensitive, for example "Aa" is not considered a palindrome here.
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Note:
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Assume the length of given string will not exceed 1,010.
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Example:
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Input:
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"abccccdd"
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Output:
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7
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Explanation:
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One longest palindrome that can be built is "dccaccd", whose length is 7.
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class Solution {
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public:
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int longestPalindrome(string s) {
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int n=s.length();
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map<char, int> freq;
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for(int i=0;i<n;i++)
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freq[s[i]]++;
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int res=0;
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for(auto it=freq.begin(); it!=freq.end(); it++){
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if(it->second>0){
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if(it->second%2==0){
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res+=it->second;
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it->second=0;
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} else {
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res+=it->second-1;
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it->second=1;
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}
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}
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}
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for(auto it=freq.begin(); it!=freq.end(); it++){
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if(it->second > 0){
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res+=1;
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break;
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}
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}
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return res;
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}
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};

‎competitive programming/leetcode/409. Longest Palindrome.cpp

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Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
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This is case sensitive, for example "Aa" is not considered a palindrome here.
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Note:
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Assume the length of given string will not exceed 1,010.
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Example:
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Input:
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"abccccdd"
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Output:
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7
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Explanation:
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One longest palindrome that can be built is "dccaccd", whose length is 7.
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class Solution {
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public:
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int longestPalindrome(string s) {
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int n=s.length();
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map<char, int> freq;
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for(int i=0;i<n;i++)
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freq[s[i]]++;
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int res=0;
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for(auto it=freq.begin(); it!=freq.end(); it++){
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if(it->second>0){
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if(it->second%2==0){
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res+=it->second;
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it->second=0;
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} else {
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res+=it->second-1;
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it->second=1;
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}
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}
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}
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for(auto it=freq.begin(); it!=freq.end(); it++){
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if(it->second > 0){
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res+=1;
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break;
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}
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}
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return res;
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}
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};

‎dynamic programming/subset_count_with_given_difference.cpp

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/*
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Given an array, we have to find the number of subsets with given difference.
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example divide the given array into 2 subsets such that S1 - S2 = diff
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where S1 = Sum of subset 1 and S2 = Sum of subset 2
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input:
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arr: 1 1 2 3
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diff: 1
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output:
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3
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*/
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#include<bits/stdc++.h>
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using namespace std;
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}
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return 0;
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}
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// We have to find count of all S1 - S2 = diff
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// We know S1 + S2 = S
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// So S1 = (diff + S) / 2

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