# Find elements in Array whose positions forms Arithmetic Progression

• Last Updated : 23 May, 2022

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++

 `// C++ code for above approach` `#include ``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);``}`

## Python3

 `## 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.`

## Javascript

 ``

Output

```1 2
2 4
3 0
5 0```

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

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