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sureshmangssureshmangs
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Practise 23-Jul-2020
1 parent e1af9ef commit 0d955d7

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‎array/merge_intervals.cpp

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/*
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Given a collection of Intervals,merge all the overlapping Intervals.
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For example:
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Given [1,3], [2,6], [8,10], [15,18],
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return [1,6], [8,10], [15,18].
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Make sure the returned intervals are sorted.
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Input:
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The first line contains an integer T, depicting total number of test cases.
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Then following T lines contains an integer N depicting the number of Intervals and next line followed by the intervals starting and ending positions 'x' and 'y' respectively.
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Output:
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Print the intervals after overlapping in sorted manner. There should be a newline at the end of output of every test case.
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Constraints:
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1 ≤ T ≤ 50
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1 ≤ N ≤ 100
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0 ≤ x ≤ y ≤ 100
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Example:
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Input
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2
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4
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1 3 2 4 6 8 9 10
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4
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6 8 1 9 2 4 4 7
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Output
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1 4 6 8 9 10
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1 9
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*/
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#include<bits/stdc++.h>
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using namespace std;
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bool comp(vector<int> i1, vector<int> i2){
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return i1[0] < i2[0];
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}
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vector<vector<int>> merge(vector<vector<int>>& intervals) {
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vector<vector<int> > res;
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if(intervals.size()<=1) return intervals;
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sort(intervals.begin(), intervals.end(), comp);
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pair<int, int> tmp;
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tmp.first=intervals[0][0];
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tmp.second=intervals[0][1];
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for(int i=1;i<intervals.size();i++){
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if(tmp.second>=intervals[i][0]){
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tmp.second=max(tmp.second, intervals[i][1]);
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} else{
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res.push_back({tmp.first, tmp.second});
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tmp.first=intervals[i][0];
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tmp.second=intervals[i][1];
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}
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}
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res.push_back({tmp.first, tmp.second}); // adding last interva;
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return res;
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}
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int main(){
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ios_base::sync_with_stdio(false);
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cin.tie(NULL);
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cout.tie(NULL);
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int t;
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cin>>t;
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while(t--){
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int n;
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cin>>n;
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vector<vector<int> > intervals, res;
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for(int i=0;i<n;i++){
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int s, e;
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cin>>s>>e;
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intervals.push_back({s, e});
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}
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res=merge(intervals);
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for(int i=0;i<res.size();i++){
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cout<<res[i][0]<<" "<<res[i][1]<<" ";
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}
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cout<<endl;
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}
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return 0;
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}

‎array/set_matrix_ones_in_O(1).cpp

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/*
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Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1.
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Input:
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The first line of input contains an integer T denoting the number of test cases.
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The first line of each test case is r and c, r is the number of rows and c is the number of columns.
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The second line of each test case contains all the elements of the matrix in a single line separated by a single space.
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Output:
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Print the modified array.
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Constraints:
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1 ≤ T ≤ 100
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1 ≤ r, c ≤ 1000
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0 ≤ A[i][j] ≤ 1
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Example:
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Input:
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3
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2 2
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1 0
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0 0
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2 3
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0 0 0
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0 0 1
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4 3
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1 0 0
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1 0 0
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1 0 0
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0 0 0
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Output:
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1 1
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1 0
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0 0 1
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1 1 1
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1 1 1
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1 1 1
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1 0 0
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Explanation:
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Testcase1: Since only first element of matrix has 1 (at index 1,1) as value, so first row and first column are modified to 1.
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*/
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#include<bits/stdc++.h>
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using namespace std;
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void setZeroes(vector<vector<int>>& matrix) {
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int m=matrix.size();
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int n=matrix[0].size();
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bool fr=false, fc=false; // check if first row or column contains a zero
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for(int i=0;i<m;i++) // col 0
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if(matrix[i][0]==1) fc=true;
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for(int j=0;j<n;j++) // row 0
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if(matrix[0][j]==1) fr=true;
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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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if(matrix[i][j]==1){
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matrix[0][j]=1;
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matrix[i][0]=1;
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}
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}
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}
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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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if(matrix[0][j]==1 || matrix[i][0]==1) matrix[i][j]=1;
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}
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}
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if(fr){
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for(int j=0;j<n;j++) matrix[0][j]=1;
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}
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if(fc){
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for(int i=0;i<m;i++) matrix[i][0]=1;
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}
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}
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int main(){
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ios_base::sync_with_stdio(false);
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cin.tie(NULL);
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cout.tie(NULL);
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int t;
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cin>>t;
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while(t--){
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int r, c;
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cin>>r>>c;
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vector<vector<int> > matrix(r, vector<int> (c, 0));
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for(int i=0;i<r;i++){
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for(int j=0;j<c;j++){
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int ch; cin>>ch;
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matrix[i][j]=ch;
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}
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}
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setZeroes(matrix);
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for(int i=0;i<r;i++){
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for(int j=0;j<c;j++){
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cout<<matrix[i][j]<<" ";
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}
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cout<<endl;
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}
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}
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return 0;
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}

‎competitive programming/leetcode/118. Pascal's Triangle.cpp

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Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
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In Pascal's triangle, each number is the sum of the two numbers directly above it.
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Example:
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Input: 5
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Output:
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[
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[1],
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[1,1],
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[1,2,1],
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[1,3,3,1],
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[1,4,6,4,1]
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]
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class Solution {
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public:
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vector<vector<int>> generate(int n) {
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vector<vector<int> > res;
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if(n==0) return res;
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int tmp[n][n];
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for(int i=0;i<n;i++){
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for(int j=0;j<=i;j++){
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if(i==j || j==0)
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tmp[i][j]=1;
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else tmp[i][j]=tmp[i-1][j-1]+tmp[i-1][j];
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}
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}
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for(int i=0;i<n;i++){
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vector<int> res1;
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for(int j=0;j<=i;j++){
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res1.push_back(tmp[i][j]);
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}
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res.push_back(res1);
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}
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return res;
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}
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};

‎competitive programming/leetcode/119. Pascal's Triangle II.cpp

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Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal's triangle.
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Note that the row index starts from 0.
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In Pascal's triangle, each number is the sum of the two numbers directly above it.
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Example:
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Input: 3
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Output: [1,3,3,1]
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Follow up:
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Could you optimize your algorithm to use only O(k) extra space?
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class Solution {
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public:
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int gcdd(int a, int b){
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if(b==0) return a;
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return gcdd(b, a%b);
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}
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int NCR(int n, int r){
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if(n==0) return 1;
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if(r==0) return 1;
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if(n-r<r) r=n-r;
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long long p=1, k=1;
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while(r){
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p*=n;
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k*=r;
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int gcd=gcdd(p, k);
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p/=gcd;
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k/=gcd;
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n--;
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r--;
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}
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return p/k;
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}
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vector<int> getRow(int n) { // works fine till n<=33
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vector<int> res;
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for(int i=0;i<=n;i++){
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int val=NCR(n, i);
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res.push_back(val);
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}
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return res;
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}
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};

‎competitive programming/leetcode/137. Single Number II.cpp

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Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
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Note:
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Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
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Example 1:
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Input: [2,2,3,2]
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Output: 3
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Example 2:
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Input: [0,1,0,1,0,1,99]
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Output: 99
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class Solution {
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public:
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int singleNumber(vector<int>& nums) {
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unordered_map<int, int> m;
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for(auto x: nums)
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m[x]++;
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for(auto it=m.begin(); it!=m.end(); it++)
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if(it->second==1) return it->first;
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return -1; // result not found
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}
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};

‎competitive programming/leetcode/2020-June-Challenge/Day-22-Single Number II.cpp

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Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
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Note:
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Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
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Example 1:
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Input: [2,2,3,2]
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Output: 3
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Example 2:
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Input: [0,1,0,1,0,1,99]
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Output: 99
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class Solution {
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public:
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int singleNumber(vector<int>& nums) {
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unordered_map<int, int> m;
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for(auto x: nums)
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m[x]++;
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for(auto it=m.begin(); it!=m.end(); it++)
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if(it->second==1) return it->first;
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return -1; // result not found
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}
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};

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