Given two arrays R[] and C[] consisting of N integers such that a matrix of dimensions N x N can be formed whose element at index (i, j) is given by (R[i] + C[j]) for all possible pairs from the two arrays. Given a matrix Queries[][] with each row representing a query of the type {A, B, X, Y}, the task for each query is to check whether there exists a path from the cell (A, B) to (X, Y) in the matrix consisting of only even numbers or not. If found to be true, print “Yes”. Otherwise, print “No”.
Note: For a path from current cell (i, j) the only possible moves are (i + 1, j), (i, j + 1), (i – 1, j), and (i, j – 1).
Examples:
Input: R[] = {6, 2, 7, 8, 3}, C[] = {3, 4, 8, 5, 1}, Queries[][] = {{2 2 1 3}, {4 2 4 3}, {5 1 3 4}}
Output: Yes Yes No
Explanation:
The matrix formed is:
{{9, 10, 14, 11, 7}
{5, 6, 10, 7, 3}
{10, 11, 15, 12, 8}
{11, 12, 16, 13, 9}
{6, 7, 11, 8, 4}}
Query1: There exist a path with even number from cell (2, 2) to (1, 3) with all cell as even number. The path is (2, 2) -> (2, 3) -> (1, 3).
Query2: There exist a path with even number from cell (4, 2) to (4, 3) with all cell as even number. The path is (4, 2) -> (4, 3).
Query3: No path exists with all elements even.Input: R[] = {1, 2, 4, 8, 5}, C[] = {2, 5, 9, 7, 6}, Queries[][] = {{2 2 1 3}, {1 4 1 3}, {5 1 3 4}}
Output: No Yes No
Explanation:
The matrix formed is:
{{3, 6, 10, 8, 7}
{4, 7, 11, 9, 8}
{6, 9, 13, 11, 10}
{10, 13, 17, 15, 14}
{7, 10, 14, 12, 11}}
Query1: No path exists with all elements even.
Query2: There exist a path with even number from cell (1, 4) to (1, 3) with all cell as even number. The path is (1, 4) -> (1, 3).
Query3: No path exists with all elements even.
Approach: The idea is to precompute the prefix sum of arrays R[] and C[] and then check for the valid path with the given conditions. Below are the steps:
- The idea is to iterate from the cell (r, c) to (r + 1, c) only if the parity of R[r] and R[r + 1] are the same because there is only one element changing.
- Similarly, do for columns from (r, c) to (r, c + 1), the parity of C and C should be the same.
- Perform the above operations for (r – 1, c) and (r, c – 1).
- Now, for each query, proceeds by checking if all elements of R[] from r = (min(A, B) to max(A, B)) have the same parity and all elements of C[] from c = (min(X, Y) to max(X, Y)) have the same parity. This can be checked in O(1) time per query by precomputing the prefix sum of parity of elements.
- If the above condition found to be true for all the cells then print “Yes” Otherwise print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find whether the path // exists with all element even or not void answerQueries(int n, int q,
vector<int>& a,
vector<int>& b,
vector<vector<int> >& Queries)
{ // Add 0 to beginning to make
// 1-based-indexing
a.insert(a.begin(), 0);
b.insert(b.begin(), 0);
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for (int i = 1; i <= n; i++) {
// Taking & to check parity
if (a[i] & 1)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for (int i = 1; i <= n; i++) {
// Taking & to check parity
if (b[i] & 1)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[][]
for (int i = 0; i < q; i++) {
int ra = Queries[i][0];
int ca = Queries[i][1];
int rb = Queries[i][2];
int cb = Queries[i][3];
int r2 = max(ra, rb), r1 = min(ra, rb);
int c2 = max(ca, cb), c1 = min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0)
|| (a[r2] - a[r1 - 1]
== r2 - r1 + 1)) {
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0)
|| (b[c2] - b[c1 - 1]
== c2 - c1 + 1)) {
// Path exists
cout << "Yes" << ' ';
continue;
}
}
// No path exists
cout << "No" << ' ';
}
} // Driver Code int main()
{ // Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
vector<int> R{ 6, 2, 7, 8, 3 };
vector<int> C{ 3, 4, 8, 5, 1 };
// Given queries[][]
vector<vector<int> > Queries{ { 2, 2, 1, 3 },
{ 4, 2, 4, 3 },
{ 5, 1, 3, 4 } };
// Function Call
answerQueries(N, Q, R, C, Queries);
return 0;
} |
Java
// Java program for the above approach import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to find whether the path // exists with all element even or not static void answerQueries(int n, int q,
int[] a, int[] b,
int[][] Queries)
{ // Add 0 to beginning to make
// 1-based-indexing
// a[0]=0;
// b[0]=0;
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((a[i] & 1) != 0)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((b[i] & 1) != 0)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[][]
for(int i = 0; i < q; i++)
{
int ra = Queries[i][0];
int ca = Queries[i][1];
int rb = Queries[i][2];
int cb = Queries[i][3];
int r2 = Math.max(ra, rb),
r1 = Math.min(ra, rb);
int c2 = Math.max(ca, cb),
c1 = Math.min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) ||
(a[r2] - a[r1 - 1] == r2 - r1 + 1))
{
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) ||
(b[c2] - b[c1 - 1] == c2 - c1 + 1))
{
// Path exists
System.out.print("Yes" + " ");
continue;
}
}
// No path exists
System.out.print("No" + " ");
}
} // Driver Code public static void main (String[] args)
throws java.lang.Exception
{ // Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
int R[] = { 0, 6, 2, 7, 8, 3 };
int C[] = { 0, 3, 4, 8, 5, 1 };
// Given queries[][]
int[][] Queries = { { 2, 2, 1, 3 },
{ 4, 2, 4, 3 },
{ 5, 1, 3, 4 }};
// Function call
answerQueries(N, Q, R, C, Queries);
}}// This code is contributed by bikram2001jha |
Python3
# Python3 program for # the above approach# Function to find whether the path# exists with all element even or notdef answerQueries(n, q, a, b, Queries):
# Add 0 to beginning to make
# 1-based-indexing
# a[0]=0;
# b[0]=0;
# Change elements based on parity
# and take prefix sum
# 1 is odd and 0 is even
# Traverse the array R
for i in range(1, n + 1):
# Taking & to check parity
if ((a[i] & 1) != 0):
a[i] = 1;
else:
a[i] = 0;
# Build prefix sum array
a[i] += a[i - 1];
# Traverse the array C
for i in range(1, n + 1):
# Taking & to check parity
if ((b[i] & 1) != 0):
b[i] = 1;
else:
b[i] = 0;
# Build prefix sum array
b[i] += b[i - 1];
# Traverse the matrix Queries
for i in range(0, q):
ra = Queries[i][0];
ca = Queries[i][1];
rb = Queries[i][2];
cb = Queries[i][3];
r2 = max(ra, rb); r1 = min(ra, rb);
c2 = max(ca, cb); c1 = min(ca, cb);
# Check if all numbers are of
# same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) or
(a[r2] - a[r1 - 1] == r2 - r1 + 1)):
# Check if all numbers are of
# same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) or
(b[c2] - b[c1 - 1] == c2 - c1 + 1)):
# Path exists
print("Yes", end = " ");
continue;
# No path exists
print("No", end = " ");
# Driver Codeif __name__ == '__main__':
# Given N, Q
N = 5;
Q = 3;
# Given array R and C
R = [0, 6, 2, 7, 8, 3];
C = [0, 3, 4, 8, 5, 1];
# Given queries
Queries = [[2, 2, 1, 3],
[4, 2, 4, 3],
[5, 1, 3, 4]];
# Function call
answerQueries(N, Q, R, C, Queries);
# This code is contributed by shikhasingrajput |
C#
// C# program for // the above approach using System;
class GFG{
// Function to find whether the path // exists with all element even or not static void answerQueries(int n, int q,
int[] a, int[] b,
int[,] Queries)
{ // Add 0 to beginning to make
// 1-based-indexing
// a[0]=0;
// b[0]=0;
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((a[i] & 1) != 0)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((b[i] & 1) != 0)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[,]
for(int i = 0; i < q; i++)
{
int ra = Queries[i, 0];
int ca = Queries[i, 1];
int rb = Queries[i, 2];
int cb = Queries[i, 3];
int r2 = Math.Max(ra, rb),
r1 = Math.Min(ra, rb);
int c2 = Math.Max(ca, cb),
c1 = Math.Min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) ||
(a[r2] - a[r1 - 1] == r2 - r1 + 1))
{
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) ||
(b[c2] - b[c1 - 1] == c2 - c1 + 1))
{
// Path exists
Console.Write("Yes" + " ");
continue;
}
}
// No path exists
Console.Write("No" + " ");
}
} // Driver Code public static void Main(String[] args)
{ // Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
int []R = {0, 6, 2, 7, 8, 3};
int []C = {0, 3, 4, 8, 5, 1};
// Given [,]queries
int[,] Queries = {{2, 2, 1, 3},
{4, 2, 4, 3},
{5, 1, 3, 4}};
// Function call
answerQueries(N, Q, R, C, Queries);
}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to implement// the above approach // Function to find whether the path // exists with all element even or not function answerQueries(n, q, a, b,Queries)
{ // Add 0 to beginning to make
// 1-based-indexing
// a[0]=0;
// b[0]=0;
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for(let i = 1; i <= n; i++)
{
// Taking & to check parity
if ((a[i] & 1) != 0)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for(let i = 1; i <= n; i++)
{
// Taking & to check parity
if ((b[i] & 1) != 0)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[][]
for(let i = 0; i < q; i++)
{
let ra = Queries[i][0];
let ca = Queries[i][1];
let rb = Queries[i][2];
let cb = Queries[i][3];
let r2 = Math.max(ra, rb),
r1 = Math.min(ra, rb);
let c2 = Math.max(ca, cb),
c1 = Math.min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) ||
(a[r2] - a[r1 - 1] == r2 - r1 + 1))
{
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) ||
(b[c2] - b[c1 - 1] == c2 - c1 + 1))
{
// Path exists
document.write("Yes" + " ");
continue;
}
}
// No path exists
document.write("No" + " ");
}
} // Driver Code // Given N, Q
let N = 5, Q = 3;
// Given array R[] and C[]
let R = [0, 6, 2, 7, 8, 3 ];
let C = [0, 3, 4, 8, 5, 1 ];
// Given queries[][]
let Queries = [[2, 2, 1, 3],
[4, 2, 4, 3 ],
[5, 1, 3, 4 ]];
// Function call
answerQueries(N, Q, R, C, Queries);
// This code is contributed by avijitmondal1998.
</script> |
Output
Yes Yes No
Time Complexity: O(N + Q)
Auxiliary Space: O(1)