Dada una array arr[] , la tarea es contar el número de veces que se actualiza el valor mínimo y máximo durante el recorrido de la array.
Ejemplos:
Entrada: arr[] = {10, 5, 20, 22}
Salida:
Número de veces que se actualiza el valor mínimo = 2
Número de veces que se actualiza el valor máximo = 3
Explicación:
Paso 1: Mínimo = 10, Máximo = 10
Paso 2: Mínimo = 5, Máximo = 10
Paso 3: Mínimo = 5, Máximo = 20
Paso 3: Mínimo = 5, Máximo = 22
Entrada: arr[] = {1, 2, 3, 4, 5}
Salida:
Número de veces que se actualiza el valor mínimo = 1
Número de veces que se actualiza el valor máximo = 5
Explicación:
Paso 1: Mínimo = 1, Máximo = 1
Paso 2: Mínimo = 1, Máximo = 2
Paso 3: Mínimo = 1, Máximo = 3
Paso 4: Mínimo = 1, Máximo = 4
Paso 5: Mínimo = 1, Máximo = 5
Enfoque: La idea es hacer un seguimiento del valor mínimo y el valor máximo. Inicialmente inicialice estos valores como el primer elemento de la array. Finalmente, itere sobre la array, y cada vez que se cambie el valor máximo o mínimo, incremente el conteo en consecuencia.
if (cur_min > arr[i])
cur_min = arr[i]
count_min++;
if (cur_max < arr[i])
cur_max = arr[i]
count_max++;
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to find the
// number of times minimum and
// maximum value updated during the
// traversal of the array
#include<bits/stdc++.h>
using namespace std;
// Function to find the number of
// times minimum and maximum value
// updated during the traversal
// of the given array
void maxUpdated(vector<int> arr)
{
int h_score = arr[0];
int l_score = arr[0];
int i = 1, j = 1;
// Increment i if new
// highest value occurs
// Increment j if new
// lowest value occurs
for (auto n : arr)
{
if (h_score < n)
{
h_score = n;
i++;
}
if (l_score > n)
{
l_score = n;
j++;
}
}
cout << "Number of times maximum value ";
cout << "updated = " << i << endl;
cout << "Number of times minimum value ";
cout << "updated = " << j << endl;
}
// Driver Code
int main()
{
vector<int> arr({10, 5, 20, 22});
maxUpdated(arr);
}
// This code is contributed by bgangwar59
Java
// Java implementation to find the
// number of times minimum and
// maximum value updated during the
// traversal of the array
public class GFG {
// Function to find the number of
// times minimum and maximum value
// updated during the traversal
// of the given array
static void maxUpdated(int[] arr)
{
int h_score = arr[0];
int l_score = arr[0];
int i = 1, j = 1;
// Increment i if new
// highest value occurs
// Increment j if new
// lowest value occurs
for (Integer n : arr) {
if (h_score < n) {
h_score = n;
i++;
}
if (l_score > n) {
l_score = n;
j++;
}
}
System.out.print(
"Number of times maximum value ");
System.out.print(
"updated = " + i + "\n");
System.out.print(
"Number of times minimum value ");
System.out.print(
"updated = " + j);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 10, 5, 20, 22 };
maxUpdated(arr);
}
}
Python
# Python implementation to count
# the number of times maximum
# and minimum value updated
# Function to find the count
# the number of times maximum
# and minimum value updated
def maximumUpdates(arr):
min = arr[0]
max = arr[0]
minCount, maxCount = 1, 1
# Update the maximum and
# minimum values during traversal
for arr in arr :
if arr>max:
maxCount+= 1
max = arr
if arr<min:
minCount+= 1;
min = arr
print("Number of times maximum ", end = "")
print("value updated = ", maxCount)
print("Number of times minimum ", end = "")
print("value updated = ", minCount)
# Driver code
if __name__ == "__main__" :
arr = [ 10, 5, 20, 22 ]
maximumUpdates(arr)
C#
// C# implementation to find the
// number of times minimum and
// maximum value updated during the
// traversal of the array
using System;
class GFG {
// Function to find the number of
// times minimum and maximum value
// updated during the traversal
// of the given array
static void maxUpdated(int[] arr)
{
int h_score = arr[0];
int l_score = arr[0];
int i = 1, j = 1;
// Increment i if new highest
// value occurs Increment j
// if new lowest value occurs
foreach(int n in arr)
{
if (h_score < n)
{
h_score = n;
i++;
}
if (l_score > n)
{
l_score = n;
j++;
}
}
Console.Write("Number of times maximum value ");
Console.Write("updated = " + i + "\n");
Console.Write("Number of times minimum value ");
Console.Write("updated = " + j);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 10, 5, 20, 22 };
maxUpdated(arr);
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// Javascript implementation to find the
// number of times minimum and
// maximum value updated during the
// traversal of the array
// Function to find the number of
// times minimum and maximum value
// updated during the traversal
// of the given array
function maxUpdated(arr)
{
let h_score = arr[0];
let l_score = arr[0];
let i = 1, j = 1;
// Increment i if new
// highest value occurs
// Increment j if new
// lowest value occurs
for (let n in arr) {
if (h_score < arr[n]) {
h_score = n;
i++;
}
if (l_score > n) {
l_score = n;
j++;
}
}
document.write(
"Number of times maximum value ");
document.write(
"updated = " + i + "<br/>");
document.write(
"Number of times minimum value ");
document.write(
"updated = " + j);
}
// Driver Code
let arr = [ 10, 5, 20, 22 ];
maxUpdated(arr);
</script>
Number of times maximum value updated = 3 Number of times minimum value updated = 2
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por utkarsh_kumar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA