Dada una array arr[] de N elementos positivos, la tarea es encontrar si es posible convertir esta array en P rogresión geométrica (GP) eliminando como máximo un elemento. En caso afirmativo, encuentre el índice de la eliminación del número que convierte la array en una progresión geométrica.
Casos especiales:
1) Si toda la array ya está en GP, devuelva cualquier índice.
2) Si no es posible convertir la array en GP, imprima «No es posible».
Ejemplos:
Input : arr[] = [2, 4, 8, 24, 16, 32] Output : 3 Number to remove is arr[3], i.e., 24 After removing 24 array will be [2, 4, 8, 16, 32] which is a GP with starting value 2 and common ratio 2. Input : arr[] = [2, 8, 30, 60] Output : Not Possible
Podemos resolver este problema manejando algunos casos especiales y luego encontrando el elemento pivote, eliminando lo que hace que la array sea una GP. Primero, verificaremos que nuestro elemento pivote sea el primer o segundo elemento, si no, el multiplicador entre ellos será la proporción común de nuestro GP, si es así, encontramos nuestra solución.
Una vez que obtengamos la relación común de GP, podemos verificar el elemento de array con esta relación, si esta relación viola algún índice, omitimos este elemento y verificamos desde el siguiente índice si es una continuación de GP anterior o no.
En el siguiente código, se implementa un método isGP que verifica que la array sea GP después de eliminar el elemento en el índice ‘índice’. Este método está escrito para el manejo de casos especiales del primer, segundo y último elemento.
Consulte el código a continuación para una mejor comprensión.
C++
// C++ program to find the element removing which
// complete array becomes a GP
#include <bits/stdc++.h>
using namespace std;
#define EPS 1e-6
// Utility method to compare two double values
bool fEqual(double a, double b)
{
return (abs(a - b) < EPS);
}
// Utility method to check, after deleting arr[ignore],
// remaining array is GP or not
bool isGP(double arr[], int N, int ignore)
{
double last = -1;
double ratio = -1;
for (int i = 0; i < N; i++)
{
// check ratio only if i is not ignore
if (i != ignore)
{
// last will be -1 first time
if (last != -1)
{
// ratio will be -1 at first time
if (ratio == -1)
ratio = (double)arr[i] / last;
// if ratio is not constant return false
else if (!fEqual(ratio, (double)arr[i] / last))
return false;
}
last = arr[i];
}
}
return true;
}
// method return value removing which array becomes GP
int makeGPbyRemovingOneElement(double arr[], int N)
{
/* solving special cases separately */
// Try removing first element
if (isGP(arr, N, 0))
return 0;
// Try removing second element
if (isGP(arr, N, 1))
return 1;
// Try removing last element
if (isGP(arr, N, N-1))
return (N-1);
/* now we know that first and second element will be
part of our GP so getting constant ratio of our GP */
double ratio = (double)arr[1]/arr[0];
for (int i = 2; i < N; i++)
{
if (!fEqual(ratio, (double)arr[i]/arr[i-1]))
{
/* At this point, we know that elements from arr[0]
to arr[i-1] are in GP. So arr[i] is the element
removing which may make GP. We check if removing
arr[i] actually makes it GP or not. */
return (isGP(arr+i-2, N-i+2, 2))? i : -1;
}
}
return -1;
}
// Driver code to test above method
int main()
{
double arr[] = {2, 4, 8, 30, 16};
int N = sizeof(arr) / sizeof(arr[0]);
int index = makeGPbyRemovingOneElement(arr, N);
if (index == -1)
cout << "Not possible\n";
else
cout << "Remove " << arr[index]
<< " to get geometric progression\n";
return 0;
}
Java
// Java program to find the element removing which
// complete array becomes a GP
import java.util.*;
class GFG {
final static double EPS = (double) 1e-6;
// Utility method to compare two double values
static boolean fEqual(double a, double b) {
return (Math.abs(a - b) < EPS);
}
// Utility method to check, after deleting arr[ignore],
// remaining array is GP or not
static boolean isGP(double[] arr, int N, int ignore) {
double last = -1;
double ratio = -1;
for (int i = 0; i < N; i++) {
// check ratio only if i is not ignore
if (i != ignore) {
// last will be -1 first time
if (last != -1) {
// ratio will be -1 at first time
if (ratio == -1)
ratio = (double) arr[i] / last;
// if ratio is not constant return false
else if (!fEqual(ratio, (double) arr[i] / last))
return false;
}
last = arr[i];
}
}
return true;
}
// method return value removing which array becomes GP
static int makeGPbyRemovingOneElement(double[] arr, int N) {
/* solving special cases separately */
// Try removing first element
if (isGP(arr, N, 0))
return 0;
// Try removing second element
if (isGP(arr, N, 1))
return 1;
// Try removing last element
if (isGP(arr, N, N - 1))
return (N - 1);
/*
* now we know that first and second element will be
* part of our GP so getting constant ratio of our GP
*/
double ratio = (double) arr[1] / arr[0];
for (int i = 2; i < N; i++) {
if (!fEqual(ratio, (double) arr[i] / arr[i - 1])) {
/*
* At this podouble, we know that elements from arr[0]
* to arr[i-1] are in GP. Soarr[i] is the element
* removing which may make GP. We check if removing
* arr[i] actually makes it GP or not.
*/
double[] temp = new double[N - i + 2];
int k = 0;
for (int j = i - 2; j < N; j++) {
temp[k++] = arr[j];
}
return (isGP(temp, N - i + 2, 2)) ? i : -1;
}
}
return -1;
}
// Driver Code
public static void main(String[] args) {
double arr[] = { 2, 4, 8, 30, 16 };
int N = arr.length;
int index = makeGPbyRemovingOneElement(arr, N);
if (index == -1)
System.out.println("Not possible");
else
System.out.println("Remove " + arr[index] +
" to get geometric progression");
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python program to find the element removing which
# complete array becomes a GP
EPS = 1e-7
# Utility method to compare two double values
def fEqual(a, b):
return abs(a - b) < EPS
# Utility method to check, after deleting arr[ignore],
# remaining array is GP or not
def isGP(arr, N, ignore):
last = -1
ratio = -1
for i in range(N):
# check ratio only if i is not ignore
if i != ignore:
# last will be -1 first time
if last != -1:
# ratio will be -1 at first time
if ratio == -1:
ratio = arr[i] / last
# if ratio is not constant return false
else if not fEqual(ratio, arr[i] / last):
return False
last = arr[i]
return True
# Method return value removing which array becomes GP
def makeGPbyRemovingOneElement(arr, N):
# Solving special cases separately
# Try removing first element
if isGP(arr, N, 0):
return 0
# Try removing second element
if isGP(arr, N, 1):
return 1
# Try removing last element
if isGP(arr, N, N - 1):
return N - 1
# now we know that first and second element will be
# part of our GP so getting constant ratio of our GP
ratio = arr[1] / arr[0]
for i in range(2, N):
if not fEqual(ratio, arr[i] / arr[i - 1]):
# At this point, we know that elements from arr[0]
# to arr[i-1] are in GP. So arr[i] is the element
# removing which may make GP. We check if removing
# arr[i] actually makes it GP or not.
return i if isGP(arr[i - 2:], N - i + 2, 2) else -1
return -1
# Driver Code
if __name__ == "__main__":
arr = [2, 4, 8, 30, 16]
N = len(arr)
index = makeGPbyRemovingOneElement(arr, N)
if index == -1:
print("Not Possible")
else:
print("Remove %d to get geometric progression" % arr[index])
# This code is contributed by
# sanjeev2552
C#
// C# program to find the element
// removing which complete array
// becomes a GP
using System;
class GFG{
static double EPS = (double)1e-6;
// Utility method to compare two
// double values
static bool fEqual(double a, double b)
{
return (Math.Abs(a - b) < EPS);
}
// Utility method to check, after
// deleting arr[ignore], remaining
// array is GP or not
static bool isGP(double[] arr, int N,
int ignore)
{
double last = -1;
double ratio = -1;
for(int i = 0; i < N; i++)
{
// Check ratio only if i is not ignore
if (i != ignore)
{
// last will be -1 first time
if (last != -1)
{
// ratio will be -1 at first time
if (ratio == -1)
ratio = (double)arr[i] / last;
// If ratio is not constant return false
else if (!fEqual(ratio,
(double) arr[i] / last))
return false;
}
last = arr[i];
}
}
return true;
}
// Method return value removing
// which array becomes GP
static int makeGPbyRemovingOneElement(double[] arr,
int N)
{
// Solving special cases separately
// Try removing first element
if (isGP(arr, N, 0))
return 0;
// Try removing second element
if (isGP(arr, N, 1))
return 1;
// Try removing last element
if (isGP(arr, N, N - 1))
return (N - 1);
// Now we know that first and second
// element will be part of our GP so
// getting constant ratio of our GP
double ratio = (double) arr[1] / arr[0];
for(int i = 2; i < N; i++)
{
if (!fEqual(ratio, (double)arr[i] /
arr[i - 1]))
{
// At this podouble, we know that
// elements from arr[0] to arr[i-1]
// are in GP. Soarr[i] is the element
// removing which may make GP. We check
// if removing arr[i] actually makes
// it GP or not.
double[] temp = new double[N - i + 2];
int k = 0;
for(int j = i - 2; j < N; j++)
{
temp[k++] = arr[j];
}
return (isGP(temp, N - i + 2, 2)) ? i : -1;
}
}
return -1;
}
// Driver Code
public static void Main(string[] args)
{
double []arr = { 2, 4, 8, 30, 16 };
int N = arr.Length;
int index = makeGPbyRemovingOneElement(arr, N);
if (index == -1)
Console.Write("Not possible");
else
Console.Write("Remove " + arr[index] +
" to get geometric progression");
}
}
// This code is contributed by rutvik_56
Javascript
// JavaScript program to find the element removing which
// complete ar4ray becomes a GP
let EPS = 1e-5;
// Utility method to compare two double values
function fEqual(a, b)
{
return Math.abs(Math.fround(a) - Math.fround(b)) < Math.fround(EPS);
}
// Utility method to check, after deleting arr[ignore],
// remaining array is GP or not
function isGP(arr, N, ignore)
{
var last = -1, ratio = -1;
for (var i = 0; i < N; i++)
{
// check ratio only if i is not ignore
if (i != ignore)
{
// last will be -1 first time
if (last != -1)
{
// ratio will be -1 at first time
if (ratio == -1)
ratio = (arr[i] / last);
// if ratio is not constant return false
else if (!(fEqual(ratio, arr[i] / last)))
return false;
}
last = arr[i];
}
}
return true;
}
// Method return value removing which array becomes GP
function makeGPbyRemovingOneElement(arr, N)
{
// Solving special cases separately
// Try removing first element
if (isGP(arr, N, 0))
return 0;
// Try removing second element
if (isGP(arr, N, 1))
return 1;
// Try removing last element
if (isGP(arr, N, N - 1))
return N - 1;
// now we know that first and second element will be
// part of our GP so getting constant ratio of our GP
var ratio = parseFloat(arr[1] / arr[0]);
for (var i = 2; i < N; i++)
{
if (!fEqual(ratio, arr[i] / arr[i - 1]))
{
// At this point, we know that elements from arr[0]
// to arr[i-1] are in GP. So arr[i] is the element
// removing which may make GP. We check if removing
// arr[i] actually makes it GP or not.
if (isGP(arr.slice(i - 2), N - i + 2, 2))
return i
return -1;
}
}
return -1
}
// Driver Code
let arr = [2, 4, 8, 30, 16];
let N = arr.length;
let index = makeGPbyRemovingOneElement(arr, N)
if (index == -1)
console.log("Not Possible")
else
console.log("Remove", arr[index], "to get geometric progression");
// This code is contributed by phasing17
Producción:
Remove 30 to get geometric progression
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA