Dada una array cuadrada de orden n*n, encuentre el máximo y el mínimo de la array dada.
Ejemplos:
Input : arr[][] = {5, 4, 9,
2, 0, 6,
3, 1, 8};
Output : Maximum = 9, Minimum = 0
Input : arr[][] = {-5, 3,
2, 4};
Output : Maximum = 4, Minimum = -5
Método ingenuo:
encontramos el máximo y el mínimo de la array por separado mediante la búsqueda lineal. El número de comparación necesario es n 2 para encontrar el mínimo y n 2 para encontrar el elemento máximo. La comparación total es igual a 2n 2 .
Comparación de pares (método eficiente):
seleccione dos elementos de la array, uno desde el comienzo de una fila de la array, otro desde el final de la misma fila de la array, compárelos y luego compare los más pequeños con el mínimo de la array y mayor de ellos al máximo de la array. Podemos ver que para dos elementos necesitamos 3 comparaciones, por lo que para recorrer toda la array necesitamos un total de 3/2 n 2 comparaciones.
Nota: Esta es una forma extendida del método 3 de Máximo Mínimo de Array.
Implementación:
C++
// C++ program for finding maximum and minimum in
// a matrix.
#include<bits/stdc++.h>
using namespace std;
#define MAX 100
// Finds maximum and minimum in arr[0..n-1][0..n-1]
// using pair wise comparisons
void maxMin(int arr[][MAX], int n)
{
int min = INT_MAX;
int max = INT_MIN;
// Traverses rows one by one
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= n/2; j++)
{
// Compare elements from beginning
// and end of current row
if (arr[i][j] > arr[i][n-j-1])
{
if (min > arr[i][n-j-1])
min = arr[i][n-j-1];
if (max< arr[i][j])
max = arr[i][j];
}
else
{
if (min > arr[i][j])
min = arr[i][j];
if (max< arr[i][n-j-1])
max = arr[i][n-j-1];
}
}
}
cout << "Maximum = " << max
<< ", Minimum = " << min;
}
/* Driver program to test above function */
int main()
{
int arr[MAX][MAX] = {5, 9, 11,
25, 0, 14,
21, 6, 4};
maxMin(arr, 3);
return 0;
}
Java
// Java program for finding maximum
// and minimum in a matrix.
class GFG
{
static final int MAX = 100;
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
static void maxMin(int arr[][], int n)
{
int min = +2147483647;
int max = -2147483648;
// Traverses rows one by one
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= n/2; j++)
{
// Compare elements from beginning
// and end of current row
if (arr[i][j] > arr[i][n - j - 1])
{
if (min > arr[i][n - j - 1])
min = arr[i][n - j - 1];
if (max< arr[i][j])
max = arr[i][j];
}
else
{
if (min > arr[i][j])
min = arr[i][j];
if (max< arr[i][n - j - 1])
max = arr[i][n - j - 1];
}
}
}
System.out.print("Maximum = "+max+
", Minimum = "+min);
}
// Driver program
public static void main (String[] args)
{
int arr[][] = {{5, 9, 11},
{25, 0, 14},
{21, 6, 4}};
maxMin(arr, 3);
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program for finding
# MAXimum and MINimum in a matrix.
MAX = 100
# Finds MAXimum and MINimum in arr[0..n-1][0..n-1]
# using pair wise comparisons
def MAXMIN(arr, n):
MIN = 10**9
MAX = -10**9
# Traverses rows one by one
for i in range(n):
for j in range(n // 2 + 1):
# Compare elements from beginning
# and end of current row
if (arr[i][j] > arr[i][n - j - 1]):
if (MIN > arr[i][n - j - 1]):
MIN = arr[i][n - j - 1]
if (MAX< arr[i][j]):
MAX = arr[i][j]
else:
if (MIN > arr[i][j]):
MIN = arr[i][j]
if (MAX< arr[i][n - j - 1]):
MAX = arr[i][n - j - 1]
print("MAXimum =", MAX, ", MINimum =", MIN)
# Driver Code
arr = [[5, 9, 11],
[25, 0, 14],
[21, 6, 4]]
MAXMIN(arr, 3)
# This code is contributed by Mohit Kumar
C#
// C# program for finding maximum
// and minimum in a matrix.
using System;
public class GFG {
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
static void maxMin(int[,] arr, int n)
{
int min = +2147483647;
int max = -2147483648;
// Traverses rows one by one
for (int i = 0; i < n; i++)
{
for (int j = 0; j <= n/2; j++)
{
// Compare elements from beginning
// and end of current row
if (arr[i,j] > arr[i,n - j - 1])
{
if (min > arr[i,n - j - 1])
min = arr[i,n - j - 1];
if (max < arr[i,j])
max = arr[i,j];
}
else
{
if (min > arr[i,j])
min = arr[i,j];
if (max < arr[i,n - j - 1])
max = arr[i,n - j - 1];
}
}
}
Console.Write("Maximum = " + max +
", Minimum = " + min);
}
// Driver code
static public void Main ()
{
int[,] arr = { {5, 9, 11},
{25, 0, 14},
{21, 6, 4} };
maxMin(arr, 3);
}
}
// This code is contributed by Shrikant13.
PHP
<?php
// PHP program for finding
// maximum and minimum in
// a matrix.
$MAX = 100;
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin($arr, $n)
{
$min = PHP_INT_MAX;
$max = PHP_INT_MIN;
// Traverses rows one by one
for ($i = 0; $i < $n; $i++)
{
for ($j = 0; $j <= $n / 2; $j++)
{
// Compare elements from beginning
// and end of current row
if ($arr[$i][$j] > $arr[$i][$n - $j - 1])
{
if ($min > $arr[$i][$n - $j - 1])
$min = $arr[$i][$n - $j - 1];
if ($max< $arr[$i][$j])
$max = $arr[$i][$j];
}
else
{
if ($min > $arr[$i][$j])
$min = $arr[$i][$j];
if ($max < $arr[$i][$n - $j - 1])
$max = $arr[$i][$n - $j - 1];
}
}
}
echo "Maximum = " , $max
,", Minimum = " , $min;
}
// Driver Code
$arr = array(array(5, 9, 11),
array(25, 0, 14),
array(21, 6, 4));
maxMin($arr, 3);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// Javascript program for finding maximum
// and minimum in a matrix.
let MAX = 100;
// Finds maximum and minimum
// in arr[0..n-1][0..n-1]
// using pair wise comparisons
function maxMin(arr,n)
{
let min = +2147483647;
let max = -2147483648;
// Traverses rows one by one
for(let i = 0; i < n; i++)
{
for(let j = 0; j <= n / 2; j++)
{
// Compare elements from beginning
// and end of current row
if (arr[i][j] > arr[i][n - j - 1])
{
if (min > arr[i][n - j - 1])
min = arr[i][n - j - 1];
if (max< arr[i][j])
max = arr[i][j];
}
else
{
if (min > arr[i][j])
min = arr[i][j];
if (max < arr[i][n - j - 1])
max = arr[i][n - j - 1];
}
}
}
document.write("Maximum = " + max +
", Minimum = " + min);
}
// Driver Code
let arr = [ [ 5, 9, 11 ],
[ 25, 0, 14 ],
[ 21, 6, 4 ] ];
maxMin(arr, 3);
// This code is contributed by sravan kumar
</script>
Maximum = 11, Minimum = 0
Complejidad temporal: O(n 2 ).
Espacio Auxiliar: O(1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA