Find elements in Array whose positions forms Arithmetic Progression
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27 Oct, 2022
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Given an array A[] of N integers. Consider an integer num such that num occurs in the array A[] and all the positions of num, sorted in increasing order forms an arithmetic progression. The task is to print all such pairs of num along with the common difference of the arithmetic progressions they form.
Examples :
Input: N = 8, A = {1, 2, 1, 3, 1, 2, 1, 5}
Output: {1, 2}, {2, 4}, {3, 0}, {5, 0}
Explanation: Positions at which 1 occurs are: 0, 2, 4, 6
It can be easily seen all the positions have same difference between consecutive elements (i.e. 2).
Hence, positions of 1 forms arithmetic progression with common difference 2.
Similarly, all positions of 2 forms arithmetic progression with common difference 4.
And, 3 and 5 are present once only.
According to the definition of arithmetic progression (or A.P.), single element sequences are also A.P. with common difference 0.Input: N = 1, A = {1}
Output: {1, 0}
Approach: The idea to solve the problem is by using Hashing.
Follow the steps to solve the problem:
- Create a map having integer as key and vector of integers as value of key.
- Traverse the array and store all positions of each element present in array into the map.
- Iterate through the map and check if all the positions of the current element in the map forms A.P.
- If so, insert that element along with the common difference into the answer.
- Else, continue to iterate through the map
Below is the implementation for the above approach:
// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all elements
// in given array whose positions forms
// arithmetic progression
void printAP(int N, int A[])
{
// Declaring a hash table(or map)
// to store positions of all elements
unordered_map<int, vector<int> > pos;
// Storing positions of
// array elements in map
for (int i = 0; i < N; i++) {
pos[A[i]].push_back(i);
}
// Declaring a map to store answer
// i.e. key - value pair of element
// with their positions forming A.P.
// and their common difference
map<int, int> ans;
// Iterating through the map "pos"
for (auto x : pos) {
// If current element is present
// only at 1 position, then simply
// insert the element in answer
// with common difference = 0
if (x.second.size() == 1) {
ans[x.first] = 0;
}
// If current element is present
// at more than one positions,
// then check if all the
// positions are at same difference
else {
bool flag = 1;
// Storing the difference of
// first two positions in
// variable "diff"
int diff = x.second[1] - x.second[0];
// Declaring "prev" to store
// previous position of
// current element
int prev = x.second[1];
//"curr" stores the
// Current position of the element
int curr;
// Iterating through all the
// positions of the current element
// and checking id they are in A.P.
for (auto it = 2;
it < x.second.size(); it++) {
curr = x.second[it];
if (curr - prev != diff) {
flag = 0;
break;
}
prev = x.second[it];
}
// If all positions of current
// element are in A.P.
// Insert it into answer with
// common difference as "diff"
if (flag == 1) {
ans[x.first] = diff;
}
}
}
// Printing the answer
for (auto it : ans) {
cout << it.first << " "
<< it.second << endl;
}
}
// Driver Code
int main()
{
int N = 8;
int A[] = { 1, 2, 1, 3, 1, 2, 1, 5 };
printAP(N, A);
}
96
1
// C++ code for above approach2
3
4
using namespace std;5
6
// Function to print all elements7
// in given array whose positions forms8
// arithmetic progression9
void printAP(int N, int A[])10
{11
12
// Declaring a hash table(or map)13
// to store positions of all elements14
unordered_map<int, vector<int> > pos;15
16
// Storing positions of17
// array elements in map18
for (int i = 0; i < N; i++) {19
pos[A[i]].push_back(i);20
}21
22
// Declaring a map to store answer23
// i.e. key - value pair of element24
// with their positions forming A.P.25
// and their common difference26
map<int, int> ans;27
28
// Iterating through the map "pos"29
for (auto x : pos) {30
31
// If current element is present32
// only at 1 position, then simply33
// insert the element in answer34
// with common difference = 035
if (x.second.size() == 1) {36
ans[x.first] = 0;37
}38
39
// If current element is present40
// at more than one positions,41
// then check if all the42
// positions are at same difference43
else {44
bool flag = 1;45
46
// Storing the difference of47
// first two positions in48
// variable "diff"49
int diff = x.second[1] - x.second[0];50
51
// Declaring "prev" to store52
// previous position of53
// current element54
int prev = x.second[1];55
56
//"curr" stores the57
// Current position of the element58
int curr;59
60
// Iterating through all the61
// positions of the current element62
// and checking id they are in A.P.63
for (auto it = 2;64
it < x.second.size(); it++) {65
curr = x.second[it];66
if (curr - prev != diff) {67
flag = 0;68
break;69
}70
prev = x.second[it];71
}72
73
// If all positions of current74
// element are in A.P.75
// Insert it into answer with76
// common difference as "diff"77
if (flag == 1) {78
ans[x.first] = diff;79
}80
}81
}82
83
// Printing the answer84
for (auto it : ans) {85
cout << it.first << " "86
<< it.second << endl;87
}88
}89
90
// Driver Code91
int main()92
{93
int N = 8;94
int A[] = { 1, 2, 1, 3, 1, 2, 1, 5 };95
printAP(N, A);96
}/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
// Function to print all elements
// in given array whose positions forms
// arithmetic progression
public static void printAP(int N, int[] A)
{
// Declaring a hash table(or map)
// to store positions of all elements
HashMap<Integer, List<Integer> > pos
= new HashMap<Integer, List<Integer> >();
// unordered_map<int, vector<int> > pos;
for (int i = 0; i < N; i++) {
pos.put(A[i], new ArrayList<Integer>());
}
// Storing positions of
// array elements in map
for (int i = 0; i < N; i++)
{
// pos(pos.get(A[i])).(i);
pos.get(A[i]).add(i);
}
// Declaring a map to store answer
// i.e. key - value pair of element
// with their positions forming A.P.
// and their common difference
Map<Integer, Integer> ans
= new HashMap<Integer, Integer>();
// map<int, int> ans;
// Iterating through the map "pos"
for (Integer x : pos.keySet()) {
Integer key = x;
List<Integer> value = pos.get(x);
// If current element is present
// only at 1 position, then simply
// insert the element in answer
// with common difference = 0
if (value.size() == 1) {
ans.put(key, 0);
}
// If current element is present
// at more than one positions,
// then check if all the
// positions are at same difference
else {
int flag = 1;
// Storing the difference of
// first two positions in
// variable "diff"
int diff = value.get(1) - value.get(0);
// Declaring "prev" to store
// previous position of
// current element
int prev = value.get(1);
//"curr" stores the
// Current position of the element
int curr;
// Iterating through all the
// positions of the current element
// and checking id they are in A.P.
for (int it = 2; it < value.size(); it++) {
curr = value.get(it);
if (curr - prev != diff) {
flag = 0;
break;
}
prev = value.get(it);
}
// If all positions of current
// element are in A.P.
// Insert it into answer with
// common difference as "diff"
if (flag == 1) {
ans.put(key, diff);
}
}
}
// Printing the answer
for (Integer it : ans.keySet()) {
int key = it;
int value = ans.get(it);
System.out.println(key + " " + value);
}
}
public static void main(String[] args)
{
int N = 8;
int[] A = { 1, 2, 1, 3, 1, 2, 1, 5 };
printAP(N, A);
}
}
// This code is contributed by akashish__
## Function to print all elements
## in given array whose positions forms
## arithmetic progression
def printAP(N, A):
## Declaring a dict
## to store positions of all elements
pos = {}
## Storing positions of
## array elements in dict
for i in range(N):
if(A[i] not in pos):
pos[A[i]] = []
pos[A[i]].append(i)
## Declaring a dict to store answer
## i.e. key - value pair of element
## with their positions forming A.P.
## and their common difference
ans = {}
## Iterating through the doct "pos"
for x in pos:
## If current element is present
## only at 1 position, then simply
## insert the element in answer
## with common difference = 0
if (len(pos[x]) == 1):
ans[x] = 0
## If current element is present
## at more than one positions,
## then check if all the
## positions are at same difference
else:
flag = 1
## Storing the difference of
## first two positions in
## variable "diff"
diff = pos[x][1]-pos[x][0]
## Declaring "prev" to store
## previous position of
## current element
prev = pos[x][1];
## "curr" stores the
## Current position of the element
curr = 0
length = len(pos[x])
## Iterating through all the
## positions of the current element
## and checking id they are in A.P.
for it in range(2, length):
curr = pos[x][it];
if (curr - prev != diff):
flag = 0;
break;
prev = curr
## If all positions of current
## element are in A.P.
## Insert it into answer with
## common difference as "diff"
if (flag == 1):
ans[x] = diff
for x in ans:
print(x, ans[x])
## Driver code
if __name__=='__main__':
N = 8
A = [1, 2, 1, 3, 1, 2, 1, 5]
printAP(N, A)
# This code is contributed by subhamgoyal2014.
/*package whatever //do not write package name here */
using System;
using System.Collections.Generic;
class GFG {
// Function to print all elements
// in given array whose positions forms
// arithmetic progression
public static void printAP(int N, int[] A)
{
// Declaring a hash table(or map)
// to store positions of all elements
Dictionary<int, List<int> > pos
= new Dictionary<int, List<int> >();
// unordered_map<int, vector<int> > pos;
for (int i = 0; i < N; i++) {
pos[A[i]] = new List<int>();
}
// Storing positions of
// array elements in map
for (int i = 0; i < N; i++)
{
// pos(pos.get(A[i])).(i);
pos[A[i]].Add(i);
}
// Declaring a map to store answer
// i.e. key - value pair of element
// with their positions forming A.P.
// and their common difference
Dictionary<int, int> ans
= new Dictionary<int, int>();
// map<int, int> ans;
// Iterating through the map "pos"
foreach (int x in pos.Keys) {
int key = x;
List<int> value = pos[x];
// If current element is present
// only at 1 position, then simply
// insert the element in answer
// with common difference = 0
if (value.Count == 1) {
ans[key] = 0;
}
// If current element is present
// at more than one positions,
// then check if all the
// positions are at same difference
else {
int flag = 1;
// Storing the difference of
// first two positions in
// variable "diff"
int diff = value[1] - value[0];
// Declaring "prev" to store
// previous position of
// current element
int prev = value[1];
//"curr" stores the
// Current position of the element
int curr;
// Iterating through all the
// positions of the current element
// and checking id they are in A.P.
for (int it = 2; it < value.Count; it++) {
curr = value[it];
if (curr - prev != diff) {
flag = 0;
break;
}
prev = value[it];
}
// If all positions of current
// element are in A.P.
// Insert it into answer with
// common difference as "diff"
if (flag == 1) {
ans[key] = diff;
}
}
}
// Printing the answer
foreach (int it in ans.Keys) {
int key = it;
int value = ans[it];
Console.WriteLine(key + " " + value);
}
}
public static void Main()
{
int N = 8;
int[] A = { 1, 2, 1, 3, 1, 2, 1, 5 };
printAP(N, A);
}
}
// This code is contributed by Saurabh Jaiswal
<script>
// JavaScript code for above approach
// Function to print all elements
// in given array whose positions forms
// arithmetic progression
const printAP = (N, A) => {
// Declaring a hash table(or map)
// to store positions of all elements
let pos = {};
// Storing positions of
// array elements in map
for (let i = 0; i < N; i++) {
if (A[i] in pos) pos[A[i]].push(i);
else pos[A[i]] = [i];
}
// Declaring a map to store answer
// i.e. key - value pair of element
// with their positions forming A.P.
// and their common difference
let ans = {};
// Iterating through the map "pos"
for (let x in pos) {
// If current element is present
// only at 1 position, then simply
// insert the element in answer
// with common difference = 0
if (pos[x].length == 1) {
ans[x] = 0;
}
// If current element is present
// at more than one positions,
// then check if all the
// positions are at same difference
else {
let flag = 1;
// Storing the difference of
// first two positions in
// variable "diff"
let diff = pos[x][1] - pos[x][0];
// Declaring "prev" to store
// previous position of
// current element
let prev = pos[x][1];
//"curr" stores the
// Current position of the element
let curr;
// Iterating through all the
// positions of the current element
// and checking id they are in A.P.
for (let it = 2; it < pos[x].length; it++) {
curr = pos[x][it];
if (curr - prev != diff) {
flag = 0;
break;
}
prev = pos[x][it];
}
// If all positions of current
// element are in A.P.
// Insert it into answer with
// common difference as "diff"
if (flag == 1) {
ans[x] = diff;
}
}
}
// Printing the answer
for (let it in ans) {
document.write(`${it} ${ans[it]}<br/>`);
}
}
// Driver Code
let N = 8;
let A = [1, 2, 1, 3, 1, 2, 1, 5];
printAP(N, A);
// This code is contributed by rakeshsahni
</script>
Output
1 2 2 4 3 0 5 0
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)
