Dada una array arr[] de tamaño N , cada entero del rango [1, N] aparece exactamente una vez excepto A que aparece dos veces y B que falta . La tarea es encontrar los números A y B.
Ejemplos:
Entrada: arr[] = {1, 2, 2, 3, 4}
Salida:
A = 2
B = 5
Entrada: arr[] = {5, 3, 4, 1, 1}
Salida:
A = 1
B = 2
Enfoque: A partir de la suma de los primeros N números naturales,
SumN = 1 + 2 + 3 + … + N = (N * (N + 1)) / 2
Y, sea la suma de todos los elementos de la array Sum . Ahora,
SumaN = Suma – A + B
A – B = Suma – SumaN …(ecuación 1)
Y de la suma de los cuadrados de los primeros N números naturales,
SumSqN = 1 2 + 2 2 + 3 2 + … + N 2 = (N * (N + 1) * (2 * n + 1)) / 6
Y, sea SumSq la suma de los cuadrados de todos los elementos del arreglo . Ahora,
SumSq = SumSqN + A 2 – B 2
SumSq – SumSqN = (A + B) * (A – B) …(ecuación 2)
Ponga el valor de (A – B) de la ecuación 1 en la ecuación 2,
SumSq – SumSqN = (A + B) * (Sum – SumN)
A + B = (SumSq – SumSqN) / (Sum – SumN) …(ecuación 3)
Resolver la ecuación 1 y la ecuación 3 dará,
B = (((SumSq – SumSqN) / (Sum – SumN)) + SumN – Sum) / 2
Y, A = Sum – SumN + B
A continuación se muestra la implementación del enfoque anterior:
C++
//C++ implementation of the approach
#include <cmath>
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
// Function to print the required numbers
void findNumbers(int arr[], int n)
{
// Sum of first n natural numbers
int sumN = (n * (n + 1)) / 2;
// Sum of squares of first n natural numbers
int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
// To store the sum and sum of squares
// of the array elements
int sum = 0, sumSq = 0, i;
for (i = 0; i < n; i++) {
sum += arr[i];
sumSq = sumSq + (pow(arr[i], 2));
}
int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
int A = sum - sumN + B;
cout << "A = " ;
cout << A << endl;
cout << "B = " ;
cout << B << endl;
}
// Driver code
int main() {
int arr[] = { 1, 2, 2, 3, 4 };
int n = sizeof(arr)/sizeof(arr[0]);
findNumbers(arr, n);
return 0;
}
Java
// Java implementation of the approach
public class GFG {
// Function to print the required numbers
static void findNumbers(int arr[], int n)
{
// Sum of first n natural numbers
int sumN = (n * (n + 1)) / 2;
// Sum of squares of first n natural numbers
int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
// To store the sum and sum of squares
// of the array elements
int sum = 0, sumSq = 0, i;
for (i = 0; i < n; i++) {
sum += arr[i];
sumSq += Math.pow(arr[i], 2);
}
int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
int A = sum - sumN + B;
System.out.println("A = " + A + "\nB = " + B);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 2, 3, 4 };
int n = arr.length;
findNumbers(arr, n);
}
}
Python3
# Python3 implementation of the approach
import math
# Function to print the required numbers
def findNumbers(arr, n):
# Sum of first n natural numbers
sumN = (n * (n + 1)) / 2;
# Sum of squares of first n natural numbers
sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
# To store the sum and sum of squares
# of the array elements
sum = 0;
sumSq = 0;
for i in range(0,n):
sum = sum + arr[i];
sumSq = sumSq + (math.pow(arr[i], 2));
B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
A = sum - sumN + B;
print("A = ",int(A)) ;
print("B = ",int(B));
# Driver code
arr = [ 1, 2, 2, 3, 4 ];
n = len(arr);
findNumbers(arr, n);
#This code is contributed by Shivi_Aggarwal
C#
// C# implementation of the approach
using System;
public class GFG {
// Function to print the required numbers
static void findNumbers(int []arr, int n)
{
// Sum of first n natural numbers
int sumN = (n * (n + 1)) / 2;
// Sum of squares of first n natural numbers
int sumSqN = (n * (n + 1) * (2 * n + 1)) / 6;
// To store the sum and sum of squares
// of the array elements
int sum = 0, sumSq = 0, i;
for (i = 0; i < n; i++) {
sum += arr[i];
sumSq += (int)Math.Pow(arr[i], 2);
}
int B = (((sumSq - sumSqN) / (sum - sumN)) + sumN - sum) / 2;
int A = sum - sumN + B;
Console.WriteLine("A = " + A + "\nB = " + B);
}
// Driver code
public static void Main()
{
int []arr = { 1, 2, 2, 3, 4 };
int n = arr.Length;
findNumbers(arr, n);
}
}
// This code is contributed by PrinciRaj1992
PHP
<?php
// PHP implementation of the approach
// Function to print the required numbers
function findNumbers($arr, $n)
{
// Sum of first n natural numbers
$sumN = ($n * ($n + 1)) / 2;
// Sum of squares of first n
// natural numbers
$sumSqN = ($n * ($n + 1) *
(2 * $n + 1)) / 6;
// To store the sum and sum of
// squares of the array elements
$sum = 0 ;
$sumSq = 0 ;
for ($i = 0; $i < $n; $i++)
{
$sum += $arr[$i];
$sumSq += pow($arr[$i], 2);
}
$B = ((($sumSq - $sumSqN) / ($sum - $sumN)) +
$sumN - $sum) / 2;
$A = $sum - $sumN + $B;
echo "A = ", $A, "\nB = ", $B;
}
// Driver code
$arr = array( 1, 2, 2, 3, 4 );
$n = sizeof($arr) ;
findNumbers($arr, $n);
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to print the required numbers
function findNumbers(arr, n)
{
// Sum of first n natural numbers
sumN = (n * (n + 1)) / 2;
// Sum of squares of first n
// natural numbers
sumSqN = (n * (n + 1) *
(2 * n + 1)) / 6;
// To store the sum and sum of
// squares of the array elements
let sum = 0 ;
let sumSq = 0 ;
for (let i = 0;i < n; i++)
{
sum += arr[i];
sumSq += Math.pow(arr[i], 2);
}
B = (((sumSq - sumSqN) / (sum - sumN)) +
sumN - sum) / 2;
A = sum - sumN + B;
document.write( "A = "+ A, "<br>B = ", B);
}
// Driver code
let arr = [ 1, 2, 2, 3, 4 ];
n = arr.length ;
findNumbers(arr, n);
// This code is contributed
// by bobby
</script>
A = 2 B = 5
Complejidad Temporal: O(N), ya que el ciclo va de 0 a (n – 1).
Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.
Publicación traducida automáticamente
Artículo escrito por AmishGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA