给定N行和M列的矩阵mat [] [] ，任务是选择任意两行(i，j)(0≤i，j≤N – 1)并构造一个新的大小为M的数组A [] ，其中A [k] = max(mat [i] [k]，mat [j] [k]) ，以使A []的最小值成为最大可能。
例子：
Input: mat[][] = {{5, 0, 3, 1, 2}, {1, 8, 9, 1, 3}, {1, 2, 3, 4, 5}, {9, 1, 0, 3, 7}, {2, 3, 0, 6, 3}, {6, 4, 1, 7, 0}}
Output: 3
Explanation:
Choose two rows i = 0 and j = 4. Then, A[] = {5, 3, 3, 6, 3} and min(A[]) is 3, which is maximum possible.
Input: mat[][] = {{2, 13, 41, 5}, {91, 11, 10, 13}, {12, 3, 28, 6}}
Output: 13
Explanation:
Choose two rows i = 0 and j = 1. Then, A[] = {91, 13, 41, 13} and min(A[]) is 13, which is maximum possible.
方法：请按照以下步骤解决问题：
- 使用INT_MAX初始化变量global_max ，该变量存储从mat [] []的任意两行构造的最小数组的最大值。
- 遍历行i和j的每种可能组合，并找到由它们构成的最小数组row_min。
- 在每次迭代中，将global_max更新为row_min及其自身的最大值。
- 完成上述步骤后，输出global_max的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
int getMaximum(int N, int M,
vector > mat)
{
// Initialize global max as INT_MIN
int global_max = INT_MIN;
// Iterate through the rows
for (int i = 0; i < N; i++) {
// Iterate through remaining rows
for (int j = i + 1; j < N; j++) {
// Initialize row_min as INT_MAX
int row_min = INT_MAX;
// Iterate through the column
// values of two rows
for (int k = 0; k < M; k++) {
// Find max of two elements
int m = max(mat[i][k],
mat[j][k]);
// Update the row_min
row_min = min(row_min, m);
}
// Update the global_max
global_max = max(global_max,
row_min);
}
}
// Print the global max
return global_max;
}
// Driver Code
int main()
{
// Given matrix mat[][]
vector > mat
= { { 5, 0, 3, 1, 2 },
{ 1, 8, 9, 1, 3 },
{ 1, 2, 3, 4, 5 },
{ 9, 1, 0, 3, 7 },
{ 2, 3, 0, 6, 3 },
{ 6, 4, 1, 7, 0 } };
// Given number of rows and columns
int N = 6, M = 5;
// Function Call
cout << getMaximum(N, M, mat);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
static int getMaximum(int N, int M, int[][] mat)
{
// Initialize global max as INT_MIN
int global_max = Integer.MIN_VALUE;
// Iterate through the rows
for(int i = 0; i < N; i++)
{
// Iterate through remaining rows
for(int j = i + 1; j < N; j++)
{
// Initialize row_min as INT_MAX
int row_min = Integer.MAX_VALUE;
// Iterate through the column
// values of two rows
for(int k = 0; k < M; k++)
{
// Find max of two elements
int m = Math.max(mat[i][k],
mat[j][k]);
// Update the row_min
row_min = Math.min(row_min, m);
}
// Update the global_max
global_max = Math.max(global_max,
row_min);
}
}
// Print the global max
return global_max;
}
// Driver Code
public static void main(String[] args)
{
// Given matrix mat[][]
int[][] mat = { { 5, 0, 3, 1, 2 },
{ 1, 8, 9, 1, 3 },
{ 1, 2, 3, 4, 5 },
{ 9, 1, 0, 3, 7 },
{ 2, 3, 0, 6, 3 },
{ 6, 4, 1, 7, 0 } };
// Given number of rows and columns
int N = 6, M = 5;
// Function Call
System.out.println(getMaximum(N, M, mat));
}
}
// This code is contributed by akhilsainiPython3
# Python3 program for the above approach
import sys
# Function to find the maximum of
# minimum of array constructed from
# any two rows of the given matrix
def getMaximum(N, M, mat):
# Initialize global max as INT_MIN
global_max = -1 * (sys.maxsize)
# Iterate through the rows
for i in range(0, N):
# Iterate through remaining rows
for j in range(i + 1, N):
# Initialize row_min as INT_MAX
row_min = sys.maxsize
# Iterate through the column
# values of two rows
for k in range(0, M):
# Find max of two elements
m = max(mat[i][k], mat[j][k])
# Update the row_min
row_min = min(row_min, m)
# Update the global_max
global_max = max(global_max,
row_min)
# Print the global max
return global_max
# Driver Code
if __name__ == '__main__':
# Given matrix mat[][]
mat = [ [ 5, 0, 3, 1, 2 ],
[ 1, 8, 9, 1, 3 ],
[ 1, 2, 3, 4, 5 ],
[ 9, 1, 0, 3, 7 ],
[ 2, 3, 0, 6, 3 ],
[ 6, 4, 1, 7, 0 ] ]
# Given number of rows and columns
N = 6
M = 5
# Function Call
print(getMaximum(N, M, mat))
# This code is contributed by akhilsainiC#
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum of
// minimum of array constructed from
// any two rows of the given matrix
static int getMaximum(int N, int M, int[,] mat)
{
// Initialize global max as INT_MIN
int global_max = int.MinValue;
// Iterate through the rows
for(int i = 0; i < N; i++)
{
// Iterate through remaining rows
for(int j = i + 1; j < N; j++)
{
// Initialize row_min as INT_MAX
int row_min = int.MaxValue;
// Iterate through the column
// values of two rows
for(int k = 0; k < M; k++)
{
// Find max of two elements
int m = Math.Max(mat[i, k],
mat[j, k]);
// Update the row_min
row_min = Math.Min(row_min, m);
}
// Update the global_max
global_max = Math.Max(global_max,
row_min);
}
}
// Print the global max
return global_max;
}
// Driver Code
public static void Main()
{
// Given matrix mat[][]
int[,] mat = { { 5, 0, 3, 1, 2 },
{ 1, 8, 9, 1, 3 },
{ 1, 2, 3, 4, 5 },
{ 9, 1, 0, 3, 7 },
{ 2, 3, 0, 6, 3 },
{ 6, 4, 1, 7, 0 } };
// Given number of rows and columns
int N = 6, M = 5;
// Function Call
Console.WriteLine(getMaximum(N, M, mat));
}
}
// This code is contributed by akhilsaini输出：
3时间复杂度： O(M * N 2 )
辅助空间： O(1)