rajan596

void solve(vector<int> &A,int l,int r,vector<int> &ans,map< pair<int,int> , long long > &mp,map< pair<int,int> , int > &index){
                
    if(r-l<=1) {
        mp[{l,r}]=0;
    }
    
    if(mp.find({l,r})!=mp.end()) return;
    
    long long cost=1e15;
    int mn=l+1;

rajan596

class SegTree{
    public:
        int n;
        int *tree;
        
        SegTree(int nn){
            n=nn;
            tree=new int[5*nn+5];
            memset(tree,0,sizeof(tree));
            build(0,n-1,0);

rajan596

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

rajan596

https://www.interviewbit.com/problems/2-sum-binary-tree/
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

rajan596

vector<int> Solution::findPerm(const string A, int B) {
        vector<int> ans;
        
        int d=0,i;
        
        for(auto x:A) {
            if(x=='D') d++;
        }
        
        i=d+1;

rajan596

int Solution::maxProduct(const vector<int> &A) {

    int mx=A[0],mn=A[0];
    int ans=A[0];
    
    for(int i=1;i<A.size();i++){
        int mx1=max(A[i],max(mx*A[i],mn*A[i]));
        int mn1=min(A[i],min(mx*A[i],mn*A[i]));
        mx=mx1;
        mn=mn1;

rajan596

#include <bits/stdc++.h>

using namespace std;

class BIT_Tree {
    private:
        int BIT[1008][1008];
    public:
        void reset(){
            memset(BIT,0,sizeof(BIT));

rajan596

/*
Array uses 0 based index and BIT uses 1 based index, so remember conversions 
*/

class BinaryIndexedTree {
  int *BIT;
  int len;
  
  BinaryIndexedTree(int size){
    BIT=new BIT[size+1];

rajan596

int Solution::repeatedNumber(const vector<int> &A) {

    int n=A.size();
    int sq=sqrt(n);
    int a[sq+5];
    
    memset(a,0,sizeof(a));

    for(auto x:A){
        int y=sqrt(x);

rajan596

int Solution::divide(int dividend, int divisor) {
    long long ans=0;
    
    int sign=1;
    
    if(dividend<0) sign=sign*(-1);
    if(divisor<0) sign=sign*(-1);
    
    long long a=abs((long)dividend);
    long long b=abs((long)divisor);