在 Matrix 中查找特定对

给定一个 nxn 整数矩阵 mat[n][n]，在 c > a 和 d > b 的所有索引选择上找到 mat(c, d) – mat(a, b) 的最大值。
例子：

Input:
mat[N][N] = {{ 1, 2, -1, -4, -20 },
             { -8, -3, 4, 2, 1 }, 
             { 3, 8, 6, 1, 3 },
             { -4, -1, 1, 7, -6 },
             { 0, -4, 10, -5, 1 }};
Output: 18
The maximum value is 18 as mat[4][2] 
- mat[1][0] = 18 has maximum difference. 

该程序应该只对矩阵进行一次遍历。即预期时间复杂度为 O(n ² )
一个简单的解决方案是应用蛮力。对于矩阵中的所有值 mat(a, b)，我们发现 mat(c, d) 具有最大值，使得 c > a 和 d > b 并不断更新迄今为止找到的最大值。我们最终返回最大值。
下面是它的实现。

C++

// A Naive method to find maximum value of mat[d][e]
// - ma[a][b] such that d > a and e > b
#include 
using namespace std;
#define N 5
 
// The function returns maximum value A(d,e) - A(a,b)
// over all choices of indexes such that both d > a
// and e > b.
int findMaxValue(int mat[][N])
{
    // stores maximum value
    int maxValue = INT_MIN;
 
    // Consider all possible pairs mat[a][b] and
    // mat[d][e]
    for (int a = 0; a < N - 1; a++)
    for (int b = 0; b < N - 1; b++)
        for (int d = a + 1; d < N; d++)
        for (int e = b + 1; e < N; e++)
            if (maxValue < (mat[d][e] - mat[a][b]))
                maxValue = mat[d][e] - mat[a][b];
 
    return maxValue;
}
 
// Driver program to test above function
int main()
{
int mat[N][N] = {
                { 1, 2, -1, -4, -20 },
                { -8, -3, 4, 2, 1 },
                { 3, 8, 6, 1, 3 },
                { -4, -1, 1, 7, -6 },
                { 0, -4, 10, -5, 1 }
            };
    cout << "Maximum Value is "
        << findMaxValue(mat);
 
    return 0;
}

Java

// A Naive method to find maximum value of mat1[d][e]
// - ma[a][b] such that d > a and e > b
import java.io.*;
import java.util.*;
  
class GFG
{
    // The function returns maximum value A(d,e) - A(a,b)
    // over all choices of indexes such that both d > a
    // and e > b.
    static int findMaxValue(int N,int mat[][])
    {
        // stores maximum value
        int maxValue = Integer.MIN_VALUE;
      
        // Consider all possible pairs mat[a][b] and
        // mat1[d][e]
        for (int a = 0; a < N - 1; a++)
          for (int b = 0; b < N - 1; b++)
             for (int d = a + 1; d < N; d++)
               for (int e = b + 1; e < N; e++)
                  if (maxValue < (mat[d][e] - mat[a][b]))
                      maxValue = mat[d][e] - mat[a][b];
      
        return maxValue;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int N = 5;
 
        int mat[][] = {
                      { 1, 2, -1, -4, -20 },
                      { -8, -3, 4, 2, 1 },
                      { 3, 8, 6, 1, 3 },
                      { -4, -1, 1, 7, -6 },
                      { 0, -4, 10, -5, 1 }
                   };
 
        System.out.print("Maximum Value is " +
                         findMaxValue(N,mat));
    }
}
  
// This code is contributed
// by Prakriti Gupta

Python 3

# A Naive method to find maximum
# value of mat[d][e] - mat[a][b]
# such that d > a and e > b
N = 5
 
# The function returns maximum
# value A(d,e) - A(a,b) over
# all choices of indexes such
# that both d > a and e > b.
def findMaxValue(mat):
     
    # stores maximum value
    maxValue = 0
 
    # Consider all possible pairs
    # mat[a][b] and mat[d][e]
    for a in range(N - 1):
        for b in range(N - 1):
            for d in range(a + 1, N):
                for e in range(b + 1, N):
                    if maxValue < int (mat[d][e] -
                                       mat[a][b]):
                        maxValue = int(mat[d][e] -
                                       mat[a][b]);
 
    return maxValue;
 
# Driver Code
mat = [[ 1, 2, -1, -4, -20 ],
       [ -8, -3, 4, 2, 1 ],
       [ 3, 8, 6, 1, 3 ],
       [ -4, -1, 1, 7, -6 ],
       [ 0, -4, 10, -5, 1 ]];
        
print("Maximum Value is " +
       str(findMaxValue(mat)))
       
# This code is contributed
# by ChitraNayal

C#

// A Naive method to find maximum
// value of mat[d][e] - mat[a][b]
// such that d > a and e > b
using System;
class GFG
{
     
    // The function returns
    // maximum value A(d,e) - A(a,b)
    // over all choices of indexes
    // such that both d > a
    // and e > b.
    static int findMaxValue(int N,
                            int [,]mat)
    {
         
        //stores maximum value
        int maxValue = int.MinValue;
     
        // Consider all possible pairs
        // mat[a][b] and mat[d][e]
        for (int a = 0; a< N - 1; a++)
        for (int b = 0; b < N - 1; b++)
            for (int d = a + 1; d < N; d++)
            for (int e = b + 1; e < N; e++)
                if (maxValue < (mat[d, e] -
                                mat[a, b]))
                    maxValue = mat[d, e] -
                               mat[a, b];
 
        return maxValue;
    }
     
    // Driver code
    public static void Main ()
    {
        int N = 5;
 
        int [,]mat = {{1, 2, -1, -4, -20},
                      {-8, -3, 4, 2, 1},
                      {3, 8, 6, 1, 3},
                      {-4, -1, 1, 7, -6},
                      {0, -4, 10, -5, 1}};
        Console.Write("Maximum Value is " +
                      findMaxValue(N,mat));
    }
}
 
// This code is contributed
// by ChitraNayal

PHP

 $a and $e > $b
$N = 5;
 
// The function returns maximum
// value A(d,e) - A(a,b) over
// all choices of indexes such
// that both $d > $a and $e > $b.
function findMaxValue(&$mat)
{
    global $N;
     
    // stores maximum value
    $maxValue = PHP_INT_MIN;
 
    // Consider all possible
    // pairs $mat[$a][$b] and
    // $mat[$d][$e]
    for ($a = 0; $a < $N - 1; $a++)
    for ($b = 0; $b < $N - 1; $b++)
        for ($d = $a + 1; $d < $N; $d++)
        for ($e = $b + 1; $e < $N; $e++)
            if ($maxValue < ($mat[$d][$e] -
                             $mat[$a][$b]))
                $maxValue = $mat[$d][$e] -
                            $mat[$a][$b];
 
    return $maxValue;
}
 
// Driver Code
$mat = array(array(1, 2, -1, -4, -20),
             array(-8, -3, 4, 2, 1),
             array(3, 8, 6, 1, 3),
             array(-4, -1, 1, 7, -6),
             array(0, -4, 10, -5, 1));
             
echo "Maximum Value is " .
       findMaxValue($mat);
 
// This code is contributed
// by ChitraNayal
?>

Javascript

// A Naive method to find maximum value of mat1[d][e]
// - ma[a][b] such that d > a and e > b   
     
    // The function returns maximum value A(d,e) - A(a,b)
    // over all choices of indexes such that both d > a
    // and e > b.
    function findMaxValue(N,mat)
    {
     
        // stores maximum value
        let maxValue = Number.MIN_VALUE;
         
        // Consider all possible pairs mat[a][b] and
        // mat1[d][e]
        for (let a = 0; a < N - 1; a++)
          for (let b = 0; b < N - 1; b++)
             for (let d = a + 1; d < N; d++)
               for (let e = b + 1; e < N; e++)
                  if (maxValue < (mat[d][e] - mat[a][b]))
                      maxValue = mat[d][e] - mat[a][b];
        
        return maxValue;
    }
     
    // Driver code
    let N = 5;
    let mat=[[ 1, 2, -1, -4, -20],[-8, -3, 4, 2, 1],[3, 8, 6, 1, 3],[ -4, -1, 1, 7, -6 ],[ 0, -4, 10, -5, 1 ]];
    document.write("Maximum Value is " +findMaxValue(N,mat));
     
    // This code is contributed by rag2127

输出：

Maximum Value is 18

上述程序在 O(n^4) 时间内运行，这与预期的 O(n^2) 时间复杂度相去甚远
一个有效的解决方案会使用额外的空间。我们对矩阵进行预处理，使得 index(i, j) 存储矩阵中从 (i, j) 到 (N-1, N-1) 的最大元素，并在此过程中不断更新迄今为止找到的最大值。我们最终返回最大值。

C++

// An efficient method to find maximum value of mat[d]
// - ma[a][b] such that c > a and d > b
#include 
using namespace std;
#define N 5
 
// The function returns maximum value A(c,d) - A(a,b)
// over all choices of indexes such that both c > a
// and d > b.
int findMaxValue(int mat[][N])
{
    //stores maximum value
    int maxValue = INT_MIN;
 
    // maxArr[i][j] stores max of elements in matrix
    // from (i, j) to (N-1, N-1)
    int maxArr[N][N];
 
    // last element of maxArr will be same's as of
    // the input matrix
    maxArr[N-1][N-1] = mat[N-1][N-1];
 
    // preprocess last row
    int maxv = mat[N-1][N-1];  // Initialize max
    for (int j = N - 2; j >= 0; j--)
    {
        if (mat[N-1][j] > maxv)
            maxv = mat[N - 1][j];
        maxArr[N-1][j] = maxv;
    }
 
    // preprocess last column
    maxv = mat[N - 1][N - 1];  // Initialize max
    for (int i = N - 2; i >= 0; i--)
    {
        if (mat[i][N - 1] > maxv)
            maxv = mat[i][N - 1];
        maxArr[i][N - 1] = maxv;
    }
 
    // preprocess rest of the matrix from bottom
    for (int i = N-2; i >= 0; i--)
    {
        for (int j = N-2; j >= 0; j--)
        {
            // Update maxValue
            if (maxArr[i+1][j+1] - mat[i][j] >
                                            maxValue)
                maxValue = maxArr[i + 1][j + 1] - mat[i][j];
 
            // set maxArr (i, j)
            maxArr[i][j] = max(mat[i][j],
                               max(maxArr[i][j + 1],
                                   maxArr[i + 1][j]) );
        }
    }
 
    return maxValue;
}
 
// Driver program to test above function
int main()
{
    int mat[N][N] = {
                      { 1, 2, -1, -4, -20 },
                      { -8, -3, 4, 2, 1 },
                      { 3, 8, 6, 1, 3 },
                      { -4, -1, 1, 7, -6 },
                      { 0, -4, 10, -5, 1 }
                    };
    cout << "Maximum Value is "
         << findMaxValue(mat);
 
    return 0;
}

Java

// An efficient method to find maximum value of mat1[d]
// - ma[a][b] such that c > a and d > b
import java.io.*;
import java.util.*;
  
class GFG
{
    // The function returns maximum value A(c,d) - A(a,b)
    // over all choices of indexes such that both c > a
    // and d > b.
    static int findMaxValue(int N,int mat[][])
    {
        //stores maximum value
        int maxValue = Integer.MIN_VALUE;
      
        // maxArr[i][j] stores max of elements in matrix
        // from (i, j) to (N-1, N-1)
        int maxArr[][] = new int[N][N];
      
        // last element of maxArr will be same's as of
        // the input matrix
        maxArr[N-1][N-1] = mat[N-1][N-1];
      
        // preprocess last row
        int maxv = mat[N-1][N-1];  // Initialize max
        for (int j = N - 2; j >= 0; j--)
        {
            if (mat[N-1][j] > maxv)
                maxv = mat[N - 1][j];
            maxArr[N-1][j] = maxv;
        }
      
        // preprocess last column
        maxv = mat[N - 1][N - 1];  // Initialize max
        for (int i = N - 2; i >= 0; i--)
        {
            if (mat[i][N - 1] > maxv)
                maxv = mat[i][N - 1];
            maxArr[i][N - 1] = maxv;
        }
      
        // preprocess rest of the matrix from bottom
        for (int i = N-2; i >= 0; i--)
        {
            for (int j = N-2; j >= 0; j--)
            {
                // Update maxValue
                if (maxArr[i+1][j+1] - mat[i][j] > maxValue)
                    maxValue = maxArr[i + 1][j + 1] - mat[i][j];
      
                // set maxArr (i, j)
                maxArr[i][j] = Math.max(mat[i][j],
                                   Math.max(maxArr[i][j + 1],
                                       maxArr[i + 1][j]) );
            }
        }
      
        return maxValue;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int N = 5;
 
        int mat[][] = {
                      { 1, 2, -1, -4, -20 },
                      { -8, -3, 4, 2, 1 },
                      { 3, 8, 6, 1, 3 },
                      { -4, -1, 1, 7, -6 },
                      { 0, -4, 10, -5, 1 }
                   };
 
        System.out.print("Maximum Value is " +
                           findMaxValue(N,mat));
    }
}
  
// Contributed by Prakriti Gupta

Python3

# An efficient method to find maximum value
# of mat[d] - ma[a][b] such that c > a and d > b
 
import sys
N = 5
 
# The function returns maximum value
# A(c,d) - A(a,b) over all choices of
# indexes such that both c > a and d > b.
def findMaxValue(mat):
 
    # stores maximum value
    maxValue = -sys.maxsize -1
 
    # maxArr[i][j] stores max of elements
    # in matrix from (i, j) to (N-1, N-1)
    maxArr = [[0 for x in range(N)]
                 for y in range(N)]
 
    # last element of maxArr will be
    # same's as of the input matrix
    maxArr[N - 1][N - 1] = mat[N - 1][N - 1]
 
    # preprocess last row
    maxv = mat[N - 1][N - 1]; # Initialize max
    for j in range (N - 2, -1, -1):
     
        if (mat[N - 1][j] > maxv):
            maxv = mat[N - 1][j]
        maxArr[N - 1][j] = maxv
     
    # preprocess last column
    maxv = mat[N - 1][N - 1] # Initialize max
    for i in range (N - 2, -1, -1):
     
        if (mat[i][N - 1] > maxv):
            maxv = mat[i][N - 1]
        maxArr[i][N - 1] = maxv
 
    # preprocess rest of the matrix
    # from bottom
    for i in range (N - 2, -1, -1):
     
        for j in range (N - 2, -1, -1):
         
            # Update maxValue
            if (maxArr[i + 1][j + 1] -
                mat[i][j] > maxValue):
                maxValue = (maxArr[i + 1][j + 1] -
                                       mat[i][j])
 
            # set maxArr (i, j)
            maxArr[i][j] = max(mat[i][j],
                           max(maxArr[i][j + 1],
                               maxArr[i + 1][j]))
         
    return maxValue
 
# Driver Code
mat = [[ 1, 2, -1, -4, -20 ],
       [-8, -3, 4, 2, 1 ],
       [ 3, 8, 6, 1, 3 ],
       [ -4, -1, 1, 7, -6] ,
       [0, -4, 10, -5, 1 ]]
                     
print ("Maximum Value is",
        findMaxValue(mat))
 
# This code is contributed by iAyushRaj

C＃

// An efficient method to find
// maximum value of mat1[d]
// - ma[a][b] such that c > a
// and d > b
using System;
class GFG  {
     
    // The function returns
    // maximum value A(c,d) - A(a,b)
    // over all choices of indexes
    // such that both c > a
    // and d > b.
    static int findMaxValue(int N, int [,]mat)
    {
         
        //stores maximum value
        int maxValue = int.MinValue;
     
        // maxArr[i][j] stores max
        // of elements in matrix
        // from (i, j) to (N-1, N-1)
        int [,]maxArr = new int[N, N];
     
        // last element of maxArr
        // will be same's as of
        // the input matrix
        maxArr[N - 1, N - 1] = mat[N - 1,N - 1];
     
        // preprocess last row
         // Initialize max
        int maxv = mat[N - 1, N - 1];
        for (int j = N - 2; j >= 0; j--)
        {
            if (mat[N - 1, j] > maxv)
                maxv = mat[N - 1, j];
            maxArr[N - 1, j] = maxv;
        }
     
        // preprocess last column
        // Initialize max
        maxv = mat[N - 1,N - 1];
        for (int i = N - 2; i >= 0; i--)
        {
            if (mat[i, N - 1] > maxv)
                maxv = mat[i,N - 1];
            maxArr[i,N - 1] = maxv;
        }
     
        // preprocess rest of the
        // matrix from bottom
        for (int i = N - 2; i >= 0; i--)
        {
            for (int j = N - 2; j >= 0; j--)
            {
                 
                // Update maxValue
                if (maxArr[i + 1,j + 1] -
                     mat[i, j] > maxValue)
                    maxValue = maxArr[i + 1,j + 1] -
                                         mat[i, j];
     
                // set maxArr (i, j)
                maxArr[i,j] = Math.Max(mat[i, j],
                              Math.Max(maxArr[i, j + 1],
                              maxArr[i + 1, j]) );
            }
        }
     
        return maxValue;
    }
     
    // Driver code
    public static void Main ()
    {
        int N = 5;
 
        int [,]mat = {{ 1, 2, -1, -4, -20 },
                      { -8, -3, 4, 2, 1 },
                      { 3, 8, 6, 1, 3 },
                      { -4, -1, 1, 7, -6 },
                      { 0, -4, 10, -5, 1 }};
        Console.Write("Maximum Value is " +
                        findMaxValue(N,mat));
    }
}
 
// This code is contributed by nitin mittal.

PHP

 a and d > b
$N = 5;
 
// The function returns maximum
// value A(c,d) - A(a,b) over
// all choices of indexes such
// that both c > a and d > b.
function findMaxValue($mat)
{
    global $N;
     
    // stores maximum value
    $maxValue = PHP_INT_MIN;
 
    // maxArr[i][j] stores max
    // of elements in matrix
    // from (i, j) to (N-1, N-1)
    $maxArr[$N][$N] = array();
 
    // last element of maxArr
    // will be same's as of
    // the input matrix
    $maxArr[$N - 1][$N - 1] = $mat[$N - 1][$N - 1];
 
    // preprocess last row
    $maxv = $mat[$N - 1][$N - 1]; // Initialize max
    for ($j = $N - 2; $j >= 0; $j--)
    {
        if ($mat[$N - 1][$j] > $maxv)
            $maxv = $mat[$N - 1][$j];
        $maxArr[$N - 1][$j] = $maxv;
    }
 
    // preprocess last column
    $maxv = $mat[$N - 1][$N - 1]; // Initialize max
    for ($i = $N - 2; $i >= 0; $i--)
    {
        if ($mat[$i][$N - 1] > $maxv)
            $maxv = $mat[$i][$N - 1];
        $maxArr[$i][$N - 1] = $maxv;
    }
 
    // preprocess rest of the
    // matrix from bottom
    for ($i = $N - 2; $i >= 0; $i--)
    {
        for ($j = $N - 2; $j >= 0; $j--)
        {
            // Update maxValue
            if ($maxArr[$i + 1][$j + 1] -
                $mat[$i][$j] > $maxValue)
                $maxValue = $maxArr[$i + 1][$j + 1] -
                            $mat[$i][$j];
 
            // set maxArr (i, j)
            $maxArr[$i][$j] = max($mat[$i][$j],
                              max($maxArr[$i][$j + 1],
                                  $maxArr[$i + 1][$j]));
        }
    }
 
    return $maxValue;
}
 
// Driver Code
$mat = array(array(1, 2, -1, -4, -20),
             array(-8, -3, 4, 2, 1),
             array(3, 8, 6, 1, 3),
             array(-4, -1, 1, 7, -6),
             array(0, -4, 10, -5, 1)
                    );
echo "Maximum Value is ".
      findMaxValue($mat);
 
// This code is contributed
// by ChitraNayal
?>

Javascript

// An efficient method to find maximum value of mat1[d]
// - ma[a][b] such that c > a and d > b
     
    // The function returns maximum value A(c,d) - A(a,b)
    // over all choices of indexes such that both c > a
    // and d > b.
    function findMaxValue(N,mat)
    {
     
        // stores maximum value
        let maxValue = Number.MIN_VALUE;
       
        // maxArr[i][j] stores max of elements in matrix
        // from (i, j) to (N-1, N-1)
        let maxArr=new Array(N);
        for(let i = 0; i < N; i++)
        {
            maxArr[i]=new Array(N);
        }
         
        // last element of maxArr will be same's as of
        // the input matrix
        maxArr[N - 1][N - 1] = mat[N - 1][N - 1];
       
        // preprocess last row
        let maxv = mat[N-1][N-1];  // Initialize max
        for (let j = N - 2; j >= 0; j--)
        {
            if (mat[N - 1][j] > maxv)
                maxv = mat[N - 1][j];
            maxArr[N - 1][j] = maxv;
        }
       
        // preprocess last column
        maxv = mat[N - 1][N - 1];  // Initialize max
        for (let i = N - 2; i >= 0; i--)
        {
            if (mat[i][N - 1] > maxv)
                maxv = mat[i][N - 1];
            maxArr[i][N - 1] = maxv;
        }
       
        // preprocess rest of the matrix from bottom
        for (let i = N-2; i >= 0; i--)
        {
            for (let j = N-2; j >= 0; j--)
            {
             
                // Update maxValue
                if (maxArr[i+1][j+1] - mat[i][j] > maxValue)
                    maxValue = maxArr[i + 1][j + 1] - mat[i][j];
       
                // set maxArr (i, j)
                maxArr[i][j] = Math.max(mat[i][j],
                                   Math.max(maxArr[i][j + 1],
                                       maxArr[i + 1][j]) );
            }
        }
       
        return maxValue;
    }
     
    // Driver code
    let N = 5;
    let mat = [[ 1, 2, -1, -4, -20 ],
           [-8, -3, 4, 2, 1 ],
           [ 3, 8, 6, 1, 3 ],
           [ -4, -1, 1, 7, -6] ,
           [0, -4, 10, -5, 1 ]];
    document.write("Maximum Value is " +
                           findMaxValue(N,mat));
           
     
    // This code is contributed by avanitrachhadiya2155

输出：

Maximum Value is 18