给定一个 N×M 矩阵和一个差值K 。任务是检查矩阵中是否存在具有给定绝对差的对。
例子：
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 100}};
K = 85
Output: YES
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8}};
K = 150
Output: NO方法：
- 初始化哈希映射以跟踪矩阵的已访问元素。
- 迭代矩阵并检查当前元素和 k 之间的差异是否已经存在于哈希映射中。如果是，则当前元素以及元素与 k 之间的差异是所需的对。
- 否则，将当前元素添加到地图中。
下面是上述方法的实现：
C++
// CPP code to check for pair with given
// difference exists in the matrix or not
#include
using namespace std;
#define N 4
#define M 4
// Function to check if a pair with given
// difference exists in the matrix
bool isPairWithDiff(int mat[N][M], int k)
{
unordered_set ump ;
// Loop to iterate over the matrix
for( int i = 0 ; i < N ; i++ )
{
for( int j =0 ; j < M ; j++ )
{
if( mat[i][j] > k )
{
int m = mat[i][j] - k ;
if( ump.find(m) != ump.end() )
{
return true ;
}
}
else
{
int m = k - mat[i][j] ;
if( ump.find(m) != ump.end() )
{
return true ;
}
}
ump.insert(mat[i][j]);
}
}
return false;
}
// Driver Code
int main()
{
// Input matrix
int mat[N][M] ={ { 5, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 100 } };
// Given difference
int k = 85;
if (isPairWithDiff(mat, k))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
} Java
// Java code to check for pair with given
// difference exists in the matrix or not
import java.util.*;
class GFG
{
static final int N = 4;
static final int M = 4;
// Function to check if a pair with given
// difference exists in the matrix
static boolean isPairWithDiff(int mat[][], int k)
{
// Store elements in a hash
HashSet s = new HashSet();
// Loop to iterate over the
// elements of the matrix
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++) {
if( mat[i][j] > k )
{
int m = mat[i][j] - k ;
if(s.contains(m))
{
return true ;
}
}
else
{
int m = k - mat[i][j] ;
if(s.contains(m))
{
return true ;
}
}
s.add(mat[i][j]);
}
}
return false;
}
// Driver Code
public static void main(String[] args)
{
// Input matrix
int mat[][] = { { 5, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 100 } };
// Given difference
int k = 85;
System.out.println(
isPairWithDiff(mat, k) == true ?
"YES" : "NO");
}
}
// This code contributed by Rajput-Ji Python3
# Python 3 program to check for pairs
# with given difference exits in the
# matrix or not
N = 4
M = 4
# Function to check if a
# pair with given
# difference exist in the matrix
def isPairWithDiff(mat, k):
# Store elements in a hash
s = set()
# Store elements in dict
for i in range(N):
for j in range(M):
if mat[i][j] > k:
m = mat[i][j] - k
if m in s:
return True
else:
m = k - mat[i][j]
if m in s:
return True
s.add(mat[i][j])
return False
# Driver Code
n, m = 4, 4
mat = [[5, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 100]]
# given difference
k = 85
if isPairWithDiff(mat, k):
print("Yes")
else:
print("No")
# This code is contributed by
# Mohit kumar 29 (IIIT gwalior)C#
// C# code to check for pair with given
// difference exists in the matrix or not
using System;
using System.Collections.Generic;
class GFG
{
static int N = 4;
static int M = 4;
// Function to check if a pair with given
// difference exists in the matrix
static Boolean isPairWithDiff(
int [,]mat, int k)
{
// Store elements in a hash
HashSet s = new HashSet();
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
if( mat[i, j] > k )
{
int m = mat[i, j] - k ;
if(s.Contains(m))
{
return true ;
}
}
else
{
int m = k - mat[i, j];
if(s.Contains(m))
{
return true ;
}
}
s.Add(mat[i, j]);
}
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
// Input matrix
int [,]mat = { { 5, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 100 } };
// Given difference
int k = 85;
Console.WriteLine(
isPairWithDiff(mat, k) == true ?
"YES" : "NO");
}
}
/* This code contributed by PrinciRaj1992 */ Javascript
输出
YES时间复杂度：O(N*M)
辅助空间： O(N*M)
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。