# Daily Coding Challenge #159

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

``````/*
Codeforces Round #677 (Div. 3) - F. Zero Remainder Sum
https://codeforces.com/contest/1433/problem/F
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vii;
typedef vector<ll> vl;
typedef vector<vl> vvl;

double EPS=1e-9;
int INF=1000000005;
long long INFF=1000000000000000005ll;
double PI=acos(-1);
int dirx[8]={ -1, 0, 0, 1, -1, -1, 1, 1 };
int diry[8]={ 0, 1, -1, 0, -1, 1, -1, 1 };
const ll MOD = 1000000007;

ll sum() { return 0; }
template<typename T, typename... Args>
T sum(T a, Args... args) { return a + sum(args...); }

#define DEBUG fprintf(stderr, "====TESTING====\n")
#define VALUE(x) cerr << "The value of " << #x << " is " << x << endl
#define OUT(x) cout << x << endl
#define OUTH(x) cout << x << " "
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define READ(x) for(auto &(z):x) cin >> z;
#define FOR(a, b, c) for (int(a)=(b); (a) < (c); ++(a))
#define FORN(a, b, c) for (int(a)=(b); (a) <= (c); ++(a))
#define FORD(a, b, c) for (int(a)=(b); (a) >= (c); --(a))
#define FORSQ(a, b, c) for (int(a)=(b); (a) * (a) <= (c); ++(a))
#define FORC(a, b, c) for (char(a)=(b); (a) <= (c); ++(a))
#define EACH(a, b) for (auto&(a) : (b))
#define REP(i, n) FOR(i, 0, n)
#define REPN(i, n) FORN(i, 1, n)
#define MAX(a, b) a=max(a, b)
#define MIN(a, b) a=min(a, b)
#define SQR(x) ((ll)(x) * (x))
#define RESET(a, b) memset(a, b, sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ALL(v) v.begin(), v.end()
#define ALLA(arr, sz) arr, arr + sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(ALL(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr, sz) sort(ALLA(arr, sz))
#define REVERSEA(arr, sz) reverse(ALLA(arr, sz))
#define PERMUTE next_permutation
#define TC(t) while (t--)
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)
#define what_is(x) cerr << #x << " is " << x << endl;

void solve() {

}

int dp[70][70][70][70]; // dp[x][y][cnt][rem]

int main()
{
FAST_INP;
//    #ifndef ONLINE_JUDGE
//    freopen("input.txt","r", stdin);
//    freopen("output.txt","w", stdout);
//    #endif

//    int tc; cin >> tc;
//    TC(tc) solve();
int n, m, k;
cin >> n >> m >> k;
vvi a(n, vi(m));
REP(i,n) REP(j,m) cin >> a[i][j];
memset(dp, -1, sizeof(dp));
function<int(int,int,int,int)> go = [&](int r, int c, int cnt, int rem) {
if(cnt > m/2) return -INT_MAX;
if(c >= m) {
r++;
c = cnt = 0;
}
if(r >= n) {
if(rem) return -INT_MAX;
return 0;
}
if(dp[r][c][cnt][rem] != -1) return dp[r][c][cnt][rem];
return dp[r][c][cnt][rem] = max(
go(r,c+1,cnt,rem),
go(r,c+1,cnt+1,(rem+a[r][c])%k) + a[r][c]
);
};
OUT(go(0,0,0,0));
return 0;
}
``````

``````/*
Minimum Depth of Binary Tree
https://leetcode.com/problems/minimum-depth-of-binary-tree/

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2
Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

The number of nodes in the tree is in the range [0, 105].
-1000 <= Node.val <= 1000
*/

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root) return 0;
int left = minDepth(root->left), right = minDepth(root->right);
return 1 + (min(left, right) ? min(left, right) : max(left, right));
}
};
``````

There are other programming solutions in the following repositories below. Star and watch for timely updates!