z0marlin

string s;
vector<string> ans;
void myFunc(int n,int o,int c){
    if(o+c==2*n){
        ans.push_back(s);
        return;
    }
    if(o<n){
        s.push_back('(');
        myFunc(n,o+1,c);

z0marlin

int pivot(const vector<int> &A){
    int l=0;
    int r=A.size()-1;
    int mid;
    while(l<r){
        mid=(r-l)/2+l;
        if(mid>l && A[mid-1]>A[mid]){
            return mid;
        }
        if(mid<r && A[mid+1]<A[mid]){

z0marlin

class Solution:
    def singleNumber(self, A):
        res=0
        count=0
        for j in range(64):
            count = 0
            for i in A :
                if 1<<j & i :
                    count+=1
            if count%3 != 0 :

z0marlin

int Solution::black(vector<string> &A) {
    int n=A.size(),m=A[0].size(),count=0,x,y;
    queue<pair<int,int>> q;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(A[i][j]=='X'){
                count++;
                q.push(make_pair(i,j));
                while(!q.empty()){
                    x=q.front().first;

z0marlin

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

z0marlin

#define SUM(i) (i >= 0 ? sum[i] : 0)
class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
    int i = 0, j = 0, len = nums.size(), min_len = INT_MAX;
    if(len == 0)
        return 0;
    int sum[len];
    sum[0] = nums[i];
    for(i = 1; i < len; i++)

z0marlin

class Solution {
public:
    bool canPartition(vector<int>& nums) {
    int sum = 0, n = nums.size();
    for (auto a : nums){
        sum += a;
    }
    if(sum&1)
        return false;

z0marlin

package main

import (
	"log"
	"os"
	"syscall"
)

const mpath = "/proc/mounts"

z0marlin

GSOC 21: Work Product Submission

Org: CNCF

Project: OpenEBS: Update mount-points and capacity of Block Devices without restarting NDM

Proposal: https://bit.ly/3D1qitD

Introduction