Given an array arr[] of N integers. The task is to check whether by arranging the elements of the array, is it possible to generate an Arithmetic Progression, Geometric Progression or Harmonic Progression. If possible print “Yes”, with the type of Progression or Else print “No”.
Examples:
Input: arr[] = {2, 16, 4, 8}
Output: Yes, A GP can be formed
Explanation:
Rearrange given array as {2, 4, 8, 16}, forms a Geometric Progression with common ratio 2.
Input: arr[] = {15, 10, 15, 5}
Output: Yes, An AP can be formed
Explanation:
Rearrange given array as {5, 10, 15, 20}, forms Arithmetic Progression with common difference 5.
Input: arr[] = { 1.0/10.0, 1.0/5.0, 1.0/15.0, 1.0/20.0 }
Output: Yes, A HP can be formed
Explanation:
Rearrange given array as { 1.0/5.0, 1.0/10.0, 1.0/15.0, 1.0/20.0 }, forms a Harmonic Progression.
Approach: The idea is to observe that elements in any of the three progressions A.P., G.P. or H.P. are somewhat related to sorted order. So, we need to first sort the given array.
- For Arithmetic Progression: Check if the difference between the consecutive elements of the sorted array are same or not. If Yes then given array element forms an Arithmetic Progression.
- For Geometric Progression: Check if the ratio of the consecutive elements of the sorted array are same or not. If Yes then given array element forms a Geometric Progression.
- For Harmonic Progression: Check if the difference between the reciprocal of all the consecutive elements of the sorted array are same or not. If Yes then given array element forms a Harmonic Progression.
Below is the implementation of the above approach:
// C++ program to check if a given // array form AP, GP or HP #include <bits/stdc++.h> using namespace std;
// Returns true if arr[0..n-1] // can form AP bool checkIsAP( double arr[], int n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
sort(arr, arr + n);
// After sorting, difference
// between consecutive elements
// must be same.
double d = arr[1] - arr[0];
// Traverse the given array and
// check if the difference
// between ith element and (i-1)th
// element is same or not
for ( int i = 2; i < n; i++) {
if (arr[i] - arr[i - 1] != d) {
return false ;
}
}
return true ;
} // Returns true if arr[0..n-1] // can form GP bool checkIsGP( double arr[], int n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
sort(arr, arr + n);
// After sorting, common ratio
// between consecutive elements
// must be same.
double r = arr[1] / arr[0];
// Traverse the given array and
// check if the common ratio
// between ith element and (i-1)th
// element is same or not
for ( int i = 2; i < n; i++) {
if (arr[i] / arr[i - 1] != r)
return false ;
}
return true ;
} // Returns true if arr[0..n-1] // can form HP bool checkIsHP( double arr[], int n)
{ // Base Case
if (n == 1) {
return true ;
}
double rec[n];
// Find reciprocal of arr[]
for ( int i = 0; i < n; i++) {
rec[i] = ((1 / arr[i]));
}
// After finding reciprocal, check if
// the reciprocal is in A. P.
// To check for A.P.
if (checkIsAP(rec, n))
return true ;
else
return false ;
} // Driver's Code int main()
{ double arr[] = { 1.0 / 5.0, 1.0 / 10.0,
1.0 / 15.0, 1.0 / 20.0 };
int n = sizeof (arr) / sizeof (arr[0]);
int flag = 0;
// Function to check AP
if (checkIsAP(arr, n)) {
cout << "Yes, An AP can be formed"
<< endl;
flag = 1;
}
// Function to check GP
if (checkIsGP(arr, n)) {
cout << "Yes, A GP can be formed"
<< endl;
flag = 1;
}
// Function to check HP
if (checkIsHP(arr, n)) {
cout << "Yes, A HP can be formed"
<< endl;
flag = 1;
}
else if (flag == 0) {
cout << "No" ;
}
return 0;
} |
// Java program to check if a given // array form AP, GP or HP import java.util.*;
class GFG{
// Returns true if arr[0..n-1] // can form AP static boolean checkIsAP( double arr[], int n)
{ // Base Case
if (n == 1 )
return true ;
// Sort array
Arrays.sort(arr);
// After sorting, difference
// between consecutive elements
// must be same.
double d = arr[ 1 ] - arr[ 0 ];
// Traverse the given array and
// check if the difference
// between ith element and (i-1)th
// element is same or not
for ( int i = 2 ; i < n; i++) {
if (arr[i] - arr[i - 1 ] != d) {
return false ;
}
}
return true ;
} // Returns true if arr[0..n-1] // can form GP static boolean checkIsGP( double arr[], int n)
{ // Base Case
if (n == 1 )
return true ;
// Sort array
Arrays.sort(arr);
// After sorting, common ratio
// between consecutive elements
// must be same.
double r = arr[ 1 ] / arr[ 0 ];
// Traverse the given array and
// check if the common ratio
// between ith element and (i-1)th
// element is same or not
for ( int i = 2 ; i < n; i++) {
if (arr[i] / arr[i - 1 ] != r)
return false ;
}
return true ;
} // Returns true if arr[0..n-1] // can form HP static boolean checkIsHP( double arr[], int n)
{ // Base Case
if (n == 1 ) {
return true ;
}
double []rec = new double [n];
// Find reciprocal of arr[]
for ( int i = 0 ; i < n; i++) {
rec[i] = (( 1 / arr[i]));
}
// After finding reciprocal, check if
// the reciprocal is in A. P.
// To check for A.P.
if (checkIsAP(rec, n))
return true ;
else
return false ;
} // Driver's Code public static void main(String[] args)
{ double arr[] = { 1.0 / 5.0 , 1.0 / 10.0 ,
1.0 / 15.0 , 1.0 / 20.0 };
int n = arr.length;
int flag = 0 ;
// Function to check AP
if (checkIsAP(arr, n)) {
System.out.print( "Yes, An AP can be formed"
+ "\n" );
flag = 1 ;
}
// Function to check GP
if (checkIsGP(arr, n)) {
System.out.print( "Yes, A GP can be formed"
+ "\n" );
flag = 1 ;
}
// Function to check HP
if (checkIsHP(arr, n)) {
System.out.print( "Yes, A HP can be formed"
+ "\n" );
flag = 1 ;
}
else if (flag == 0 ) {
System.out.print( "No" );
}
} } // This code is contributed by PrinciRaj1992 |
# Python3 program to check if a # given array form AP, GP or HP # Returns true if arr[0..n-1] # can form AP def checkIsAP(arr, n):
# Base Case
if (n = = 1 ):
return True
# Sort array
arr.sort();
# After sorting, difference
# between consecutive elements
# must be same.
d = arr[ 1 ] - arr[ 0 ]
# Traverse the given array and
# check if the difference
# between ith element and (i-1)th
# element is same or not
for i in range ( 2 , n):
if (arr[i] - arr[i - 1 ] ! = d):
return False
return True
# Returns true if arr[0..n-1] # can form GP def checkIsGP(arr, n):
# Base Case
if (n = = 1 ):
return True
# Sort array
arr.sort()
# After sorting, common ratio
# between consecutive elements
# must be same.
r = arr[ 1 ] / arr[ 0 ]
# Traverse the given array and
# check if the common ratio
# between ith element and (i-1)th
# element is same or not
for i in range ( 2 , n):
if (arr[i] / arr[i - 1 ] ! = r):
return False
return True
# Returns true if arr[0..n-1] # can form HP def checkIsHP(arr, n):
# Base Case
if (n = = 1 ):
return True
rec = []
# Find reciprocal of arr[]
for i in range ( 0 , n):
rec.append(( 1 / arr[i]))
# After finding reciprocal, check
# if the reciprocal is in A. P.
# To check for A.P.
if (checkIsAP(rec, n)):
return True
else :
return False
# Driver Code arr = [ 1.0 / 5.0 , 1.0 / 10.0 ,
1.0 / 15.0 , 1.0 / 20.0 ]
n = len (arr)
flag = 0
# Function to check AP if (checkIsAP(arr, n)):
print ( "Yes, An AP can be formed" , end = '\n' )
flag = 1
# Function to check GP if (checkIsGP(arr, n)):
print ( "Yes, A GP can be formed" , end = '\n' )
flag = 1
# Function to check HP if (checkIsHP(arr, n)):
print ( "Yes, A HP can be formed" , end = '\n' )
flag = 1
elif (flag = = 0 ):
print ( "No" , end = '\n' )
# This code is contributed by Pratik |
// C# program to check if a given // array form AP, GP or HP using System;
class GFG{
// Returns true if arr[0..n-1] // can form AP static bool checkIsAP( double []arr, int n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
Array.Sort(arr);
// After sorting, difference
// between consecutive elements
// must be same.
double d = arr[1] - arr[0];
// Traverse the given array and
// check if the difference
// between ith element and (i-1)th
// element is same or not
for ( int i = 2; i < n; i++) {
if (arr[i] - arr[i - 1] != d) {
return false ;
}
}
return true ;
} // Returns true if arr[0..n-1] // can form GP static bool checkIsGP( double []arr, int n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
Array.Sort(arr);
// After sorting, common ratio
// between consecutive elements
// must be same.
double r = arr[1] / arr[0];
// Traverse the given array and
// check if the common ratio
// between ith element and (i-1)th
// element is same or not
for ( int i = 2; i < n; i++) {
if (arr[i] / arr[i - 1] != r)
return false ;
}
return true ;
} // Returns true if arr[0..n-1] // can form HP static bool checkIsHP( double []arr, int n)
{ // Base Case
if (n == 1) {
return true ;
}
double []rec = new double [n];
// Find reciprocal of []arr
for ( int i = 0; i < n; i++) {
rec[i] = ((1 / arr[i]));
}
// After finding reciprocal, check if
// the reciprocal is in A. P.
// To check for A.P.
if (checkIsAP(rec, n))
return true ;
else
return false ;
} // Driver's Code public static void Main(String[] args)
{ double []arr = { 1.0 / 5.0, 1.0 / 10.0,
1.0 / 15.0, 1.0 / 20.0 };
int n = arr.Length;
int flag = 0;
// Function to check AP
if (checkIsAP(arr, n)) {
Console.Write( "Yes, An AP can be formed"
+ "\n" );
flag = 1;
}
// Function to check GP
if (checkIsGP(arr, n)) {
Console.Write( "Yes, A GP can be formed"
+ "\n" );
flag = 1;
}
// Function to check HP
if (checkIsHP(arr, n)) {
Console.Write( "Yes, A HP can be formed"
+ "\n" );
flag = 1;
}
else if (flag == 0) {
Console.Write( "No" );
}
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript program to check if a given // array form AP, GP or HP // Returns true if arr[0..n-1] // can form AP function checkIsAP(arr, n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
arr.sort( function (a,b){ return a-b;});
// After sorting, difference
// between consecutive elements
// must be same.
var d = arr[1] - arr[0];
// Traverse the given array and
// check if the difference
// between ith element and (i-1)th
// element is same or not
for ( var i = 2; i < n; i++) {
if (arr[i] - arr[i - 1] != d) {
return false ;
}
}
return true ;
} // Returns true if arr[0..n-1] // can form GP function checkIsGP(arr, n)
{ // Base Case
if (n == 1)
return true ;
// Sort array
arr.sort();
// After sorting, common ratio
// between consecutive elements
// must be same.
var r = arr[1] / arr[0];
// Traverse the given array and
// check if the common ratio
// between ith element and (i-1)th
// element is same or not
for ( var i = 2; i < n; i++) {
if (arr[i] / arr[i - 1] != r)
return false ;
}
return true ;
} // Returns true if arr[0..n-1] // can form HP function checkIsHP(arr, n)
{ // Base Case
if (n == 1) {
return true ;
}
var rec = Array(n).fill(0);
// Find reciprocal of arr[]
for ( var i = 0; i < n; i++) {
rec[i] = ((1 / arr[i]));
}
// After finding reciprocal, check if
// the reciprocal is in A. P.
// To check for A.P.
if (checkIsAP(rec, n))
return true ;
else
return false ;
} // Driver's Code var arr = [ 1.0 / 5.0, 1.0 / 10.0,
1.0 / 15.0, 1.0 / 20.0 ];
var n = arr.length;
var flag = 0;
// Function to check AP if (checkIsAP(arr, n)) {
document.write( "Yes, An AP can be formed" );
flag = 1;
} // Function to check GP if (checkIsGP(arr, n)) {
document.write( "Yes, A GP can be formed" );
flag = 1;
} // Function to check HP if (checkIsHP(arr, n)) {
document.write( "Yes, A HP can be formed" );
flag = 1;
} else if (flag == 0) {
document.write( "No" );
} </script> |
Yes, A HP can be formed
Time Complexity: O(N*log N)
Auxiliary Space: O(N)