Grid Challenge


Problem Statement :


Given a square grid of characters in the range ascii[a-z], rearrange elements of each row alphabetically, ascending. Determine if the columns are also in ascending alphabetical order, top to bottom. Return YES if they are or NO if they are not.

Example

The grid is illustrated below.

a b c
a d e
e f g

The rows are already in alphabetical order. The columns a a e, b d f and c e g are also in alphabetical order, so the answer would be YES. Only elements within the same row can be rearranged. They cannot be moved to a different row.


Function Description

Complete the gridChallenge function in the editor below.

gridChallenge has the following parameter(s):

string grid[n]: an array of strings
Returns

string: either YES or NO

Input Format

The first line contains , the number of testcases.

Each of the next  sets of lines are described as follows:
- The first line contains , the number of rows and columns in the grid.
- The next  lines contains a string of length 


Output Format

For each test case, on a separate line print YES if it is possible to rearrange the grid alphabetically ascending in both its rows and columns, or NO otherwise.


Solution :



title-img 


                            Solution in C :

In  C :






#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
typedef long long ll;
int main() {
    char m[105][105];
    ll t,n,i,j,f,cnt[26],k;
    ll a[105][105];
    scanf("%lld",&t);
    while(t--)
        {
        scanf("%lld",&n);
        for(i=0;i<n;i++)
            {
            scanf("%s",m[i]);
        }
        for(i=0;i<n;i++)
            {
            memset(cnt,0,sizeof(cnt));
            for(j=0;j<n;j++)
                {
                cnt[m[i][j]-'a']++;;
            }
            j=0;
            for(k=0;k<26;k++)
                {
                while(cnt[k]>0)
                    {
                    a[i][j]=k;
                    j++;
                    cnt[k]--;
                }
            }
        }
        f=0;
        for(j=0;j<n&&f==0;j++)
            {
            for(i=0;(i+1)<n&&f==0;i++)
                {
                if(a[i][j]>a[i+1][j])
                    {
                    f=1;
                }
            }
        }
        if(f==0)
            {
            printf("YES\n");
        }
        else
            {
            printf("NO\n");
        }
    }
    return 0;
}
                        


                        Solution in C++ :

In  C++  :







#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_DEPRECATE

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>
#include <iomanip>
#include <fstream>
#include <iterator>
#include <bitset>

using namespace std;

typedef long long ll;
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

#define FOR(i,n) for(int i = 0; i < (n); i++)
#define sz(c) ((int)(c).size())
#define ten(x) ((int)1e##x)
#define tenll(x) ((ll)1e##x)
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

void solve(){
	int n; cin >> n;
	vector<string> v(n);
	FOR(i, n) cin >> v[i];
	FOR(i, n) sort(v[i].begin(), v[i].end());
	bool b = true;
	FOR(i, n){
		string x;
		FOR(j, n) x.push_back(v[j][i]);
		string y = x;
		sort(y.begin(), y.end());
		if (x != y) {
			b = false;
		}
	}
	puts(b ? "YES" : "NO");
}

int main(){
	int t; cin >> t;
	while (t--) {
		solve();
	}
	return 0;
}
                    


                        Solution in Java :

In  Java :







import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.math.BigInteger;
import java.io.InputStream;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		OutputWriter out = new OutputWriter(outputStream);
		GridChallenge solver = new GridChallenge();
		int testCount = Integer.parseInt(in.next());
		for (int i = 1; i <= testCount; i++)
			solver.solve(i, in, out);
		out.close();
	}
}

class GridChallenge {
    public void solve(int testNumber, InputReader in, OutputWriter out) {
        int n = in.readInt();
        char[][] g = new char[n][n];
        for (int i = 0; i < n; i++) {
            g[i] = in.readString().toCharArray();
            Arrays.sort(g[i]);
        }
        boolean ok = true;
        for (int j = 0; j < n && ok; j++) {
            for (int i = 0; i < n - 1; i++) {
                if (g[i + 1][j] < g[i][j]) {
                    ok = false;
                    break;
                }
            }
        }
        if (ok) {
            out.printLine("YES");
        }
        else {
            out.printLine("NO");
        }
    }
}

class InputReader {

    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    private SpaceCharFilter filter;

    public InputReader(InputStream stream) {
        this.stream = stream;
    }

    public int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    public int readInt() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int res = 0;
        do {
            if (c < '0' || c > '9')
                throw new InputMismatchException();
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }

    public String readString() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    public boolean isSpaceChar(int c) {
        if (filter != null)
            return filter.isSpaceChar(c);
        return isWhitespace(c);
    }

    public static boolean isWhitespace(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public String next() {
        return readString();
    }

    public interface SpaceCharFilter {
        public boolean isSpaceChar(int ch);
    }
}

class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public void print(Object...objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }

    public void printLine(Object...objects) {
        print(objects);
        writer.println();
    }

    public void close() {
        writer.close();
    }

}
                    


                        Solution in Python : 
                            
In  Python3 :





def sort_string(string):
    return "".join(sorted(string))
for _ in range(int(input())):
    n = int(input())
    string = []
    for i in range(n):
        s = input()
        string.append(sort_string(s))
    value = True
    for j in range(n):
        for i in range(n-1):
            if string[i][j] > string[i+1][j]:
                value = False
                break
        if not value:
            break
    
    if value:
        print('YES')
    else:
        print('NO')


View More Similar Problems

Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →

Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →

QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →

Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

View Solution →