Dada una array de enteros, la tarea es reemplazar cada elemento con el elemento más pequeño en su lado izquierdo.
Nota: Reemplace el primer elemento con -1 ya que no tiene ningún elemento a la izquierda.
Ejemplos:
Input: arr[] = {4, 5, 2, 1, 7, 6}
Output: -1 4 4 2 1 1
Since, 4 has no element in its left, so replace it by -1.
For 5, 4 is the smallest element in its left.
For 2, 4 is the smallest element in its left.
For 1, 2 is the smallest element in its left.
For 7, 1 is the smallest element in its left.
For 6, 1 is the smallest element in its left.
Input: arr[] = {3, 2, 5, 7, 1}
Output: -1 3 2 2 2
Acercarse:
- Mantenga una variable min_ele que almacenará el elemento más pequeño.
- Inicialmente, min_ele será igual al elemento en el índice 0.
- Primero reemplace arr[0] con -1 y luego recorra la array
- Y luego reemplace el elemento con el valor min_ele y actualice min_ele si el elemento es más pequeño que min_ele.
- Imprime la array modificada.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to Replace every
// element with the smaller element
// on its left side
#include <bits/stdc++.h>
using namespace std;
// Function to replace the elements
void ReplaceElements(int arr[], int n)
{
// MIN value initialised
// to element at 0th index
int min_ele = arr[0];
arr[0] = -1;
for (int i = 1; i < n; ++i) {
// If min_ele is smaller than arr[i]
// then just replace arr[i] with min_ele
if (min_ele < arr[i])
arr[i] = min_ele;
// Else if update the min_ele also
else if (min_ele >= arr[i]) {
int temp = arr[i];
arr[i] = min_ele;
min_ele = temp;
}
}
}
// Driver code
int main()
{
int arr[] = { 4, 5, 2, 1, 7, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
// Replace the elements
// with the smaller element
// on its left side
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; ++i)
cout << arr[i] << " ";
return 0;
}
Java
// Java program to Replace every
// element with the smaller element
// on its left side
class GFG {
// Function to replace the elements
static void ReplaceElements(int arr[], int n) {
// MIN value initialised
// to element at 0th index
int min_ele = arr[0];
arr[0] = -1;
for (int i = 1; i < n; ++i) {
// If min_ele is smaller than arr[i]
// then just replace arr[i] with min_ele
if (min_ele < arr[i]) {
arr[i] = min_ele;
} // Else if update the min_ele also
else if (min_ele >= arr[i]) {
int temp = arr[i];
arr[i] = min_ele;
min_ele = temp;
}
}
}
// Driver code
public static void main(String[] args) {
int arr[] = {4, 5, 2, 1, 7, 6};
int n = arr.length;
// Replace the elements
// with the smaller element
// on its left side
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; ++i) {
System.out.print(arr[i] + " ");
}
}
}
// This code is contributed by Rajput-JI
Python3
# Python3 program to Replace every # element with the smaller element # on its left side # Function to replace the elements def ReplaceElements(arr, n): # MIN value initialised # to element at 0th index min_ele = arr[0] arr[0] = -1 for i in range(1, n): # If min_ele is smaller than arr[i] # then just replace arr[i] with min_ele if (min_ele < arr[i]): arr[i] = min_ele # Else if update the min_ele also elif (min_ele >= arr[i]) : temp = arr[i] arr[i] = min_ele min_ele = temp # Driver code if __name__ == "__main__": arr = [ 4, 5, 2, 1, 7, 6 ] n = len (arr) # Replace the elements # with the smaller element # on its left side ReplaceElements(arr, n) # Print the modified array for i in range( n): print (arr[i], end = " ") # This code is contributed # by ChitraNayal
C#
// C# program to Replace every
// element with the smaller element
// on its left side
using System;
public class GFG {
// Function to replace the elements
static void ReplaceElements(int []arr, int n) {
// MIN value initialised
// to element at 0th index
int min_ele = arr[0];
arr[0] = -1;
for (int i = 1; i < n; ++i) {
// If min_ele is smaller than arr[i]
// then just replace arr[i] with min_ele
if (min_ele < arr[i]) {
arr[i] = min_ele;
} // Else if update the min_ele also
else if (min_ele >= arr[i]) {
int temp = arr[i];
arr[i] = min_ele;
min_ele = temp;
}
}
}
// Driver code
public static void Main() {
int []arr = {4, 5, 2, 1, 7, 6};
int n = arr.Length;
// Replace the elements
// with the smaller element
// on its left side
ReplaceElements(arr, n);
// Print the modified array
for (int i = 0; i < n; ++i) {
Console.Write(arr[i] + " ");
}
}
}
// This code is contributed by Rajput-JI
PHP
<?php
// PHP program to Replace every
// element with the smaller element
// on its left side
// Function to replace the elements
function ReplaceElements($arr, $n)
{
// MIN value initialised
// to element at 0th index
$min_ele = $arr[0];
$arr[0] = -1;
for ($i = 1; $i < $n; ++$i)
{
// If min_ele is smaller than arr[i]
// then just replace arr[i] with min_ele
if ($min_ele < $arr[$i])
$arr[$i] = $min_ele;
// Else if update the min_ele also
else if ($min_ele >= $arr[$i])
{
$temp = $arr[$i];
$arr[$i] = $min_ele;
$min_ele = $temp;
}
}
return $arr;
}
// Driver code
$arr = array(4, 5, 2, 1, 7, 6);
$n = sizeof($arr);
// Replace the elements
// with the smaller element
// on its left side
$arr1 = ReplaceElements($arr, $n);
// Print the modified array
for ($i = 0; $i < $n; ++$i)
echo $arr1[$i] . " ";
// This code is contributed
// by Akanksha Rai
?>
Javascript
<script>
// JavaScript program to Replace every
// element with the smaller element
// on its left side
// Function to replace the elements
function ReplaceElements(arr, n)
{
// MIN value initialised
// to element at 0th index
var min_ele = arr[0];
arr[0] = -1;
for (var i = 1; i < n; ++i)
{
// If min_ele is smaller than arr[i]
// then just replace arr[i] with min_ele
if (min_ele < arr[i]) arr[i] = min_ele;
// Else if update the min_ele also
else if (min_ele >= arr[i])
{
var temp = arr[i];
arr[i] = min_ele;
min_ele = temp;
}
}
}
// Driver code
var arr = [4, 5, 2, 1, 7, 6];
var n = arr.length;
// Replace the elements
// with the smaller element
// on its left side
ReplaceElements(arr, n);
// Print the modified array
for (var i = 0; i < n; ++i) document.write(arr[i] + " ");
// This code is contributed by rdtank.
</script>
Producción:
-1 4 4 2 1 1
Complejidad de tiempo: O(N)
Publicación traducida automáticamente
Artículo escrito por imdhruvgupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA