Given an array that represents elements of arithmetic progression in order. One element is missing in the progression, find the missing number.

**Examples: **

Input: arr[] = {2, 4, 8, 10, 12, 14} Output: 6 Input: arr[] = {1, 6, 11, 16, 21, 31}; Output: 26

A **Simple Solution** is to linearly traverse the array and find the missing number. Time complexity of this solution is O(n).

We can solve this problem **in O(Logn) time** using Binary Search. The idea is to go to the middle element. Check if the difference between middle and next to middle is equal to diff or not, if not then the missing element lies between mid and mid+1. If the middle element is equal to n/2^{th} term in Arithmetic Series (Let n be the number of elements in input array), then missing element lies in right half. Else element lies in left half.

Following is implementation of above idea.

## C

`// A C program to find the missing number in a given ` `// arithmetic progression ` `#include <stdio.h> ` `#include <limits.h> ` ` ` `// A binary search based recursive function that returns ` `// the missing element in arithmetic progression ` `int` `findMissingUtil(` `int` `arr[], ` `int` `low, ` `int` `high, ` `int` `diff) ` `{ ` ` ` `// There must be two elements to find the missing ` ` ` `if` `(high <= low) ` ` ` `return` `INT_MAX; ` ` ` ` ` `// Find index of middle element ` ` ` `int` `mid = low + (high - low)/2; ` ` ` ` ` `// The element just after the middle element is missing. ` ` ` `// The arr[mid+1] must exist, because we return when ` ` ` `// (low == high) and take floor of (high-low)/2 ` ` ` `if` `(arr[mid+1] - arr[mid] != diff) ` ` ` `return` `(arr[mid] + diff); ` ` ` ` ` `// The element just before mid is missing ` ` ` `if` `(mid > 0 && arr[mid] - arr[mid-1] != diff) ` ` ` `return` `(arr[mid-1] + diff); ` ` ` ` ` `// If the elements till mid follow AP, then recur ` ` ` `// for right half ` ` ` `if` `(arr[mid] == arr[0] + mid*diff) ` ` ` `return` `findMissingUtil(arr, mid+1, high, diff); ` ` ` ` ` `// Else recur for left half ` ` ` `return` `findMissingUtil(arr, low, mid-1, diff); ` `} ` ` ` `// The function uses findMissingUtil() to find the missing ` `// element in AP. It assumes that there is exactly one missing ` `// element and may give incorrect result when there is no missing ` `// element or more than one missing elements. ` `// This function also assumes that the difference in AP is an ` `// integer. ` `int` `findMissing(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// If exactly one element is missing, then we can find ` ` ` `// difference of arithmetic progression using following ` ` ` `// formula. Example, 2, 4, 6, 10, diff = (10-2)/4 = 2. ` ` ` `// The assumption in formula is that the difference is ` ` ` `// an integer. ` ` ` `int` `diff = (arr[n-1] - arr[0])/n; ` ` ` ` ` `// Binary search for the missing number using above ` ` ` `// calculated diff ` ` ` `return` `findMissingUtil(arr, 0, n-1, diff); ` `} ` ` ` `/* Driver program to check above functions */` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 4, 8, 10, 12, 14}; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); ` ` ` `printf` `(` `"The missing element is %d"` `, findMissing(arr, n)); ` ` ` `return` `0; ` `} ` |

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## Java

`// A Java program to find ` `// the missing number in ` `// a given arithmetic ` `// progression ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// A binary search based ` `// recursive function that ` `// returns the missing ` `// element in arithmetic ` `// progression ` `static` `int` `findMissingUtil(` `int` `arr[], ` `int` `low, ` ` ` `int` `high, ` `int` `diff) ` `{ ` ` ` `// There must be two elements ` ` ` `// to find the missing ` ` ` `if` `(high <= low) ` ` ` `return` `Integer.MAX_VALUE; ` ` ` ` ` `// Find index of ` ` ` `// middle element ` ` ` `int` `mid = low + (high - low) / ` `2` `; ` ` ` ` ` `// The element just after the ` ` ` `// middle element is missing. ` ` ` `// The arr[mid+1] must exist, ` ` ` `// because we return when ` ` ` `// (low == high) and take ` ` ` `// floor of (high-low)/2 ` ` ` `if` `(arr[mid + ` `1` `] - arr[mid] != diff) ` ` ` `return` `(arr[mid] + diff); ` ` ` ` ` `// The element just ` ` ` `// before mid is missing ` ` ` `if` `(mid > ` `0` `&& arr[mid] - ` ` ` `arr[mid - ` `1` `] != diff) ` ` ` `return` `(arr[mid - ` `1` `] + diff); ` ` ` ` ` `// If the elements till mid follow ` ` ` `// AP, then recur for right half ` ` ` `if` `(arr[mid] == arr[` `0` `] + mid * diff) ` ` ` `return` `findMissingUtil(arr, mid + ` `1` `, ` ` ` `high, diff); ` ` ` ` ` `// Else recur for left half ` ` ` `return` `findMissingUtil(arr, low, mid - ` `1` `, diff); ` `} ` ` ` `// The function uses findMissingUtil() ` `// to find the missing element in AP. ` `// It assumes that there is exactly ` `// one missing element and may give ` `// incorrect result when there is no ` `// missing element or more than one ` `// missing elements. This function also ` `// assumes that the difference in AP is ` `// an integer. ` `static` `int` `findMissing(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `// If exactly one element is missing, ` ` ` `// then we can find difference of ` ` ` `// arithmetic progression using ` ` ` `// following formula. Example, 2, 4, ` ` ` `// 6, 10, diff = (10-2)/4 = 2. ` ` ` `// The assumption in formula is that ` ` ` `// the difference is an integer. ` ` ` `int` `diff = (arr[n - ` `1` `] - arr[` `0` `]) / n; ` ` ` ` ` `// Binary search for the missing ` ` ` `// number using above calculated diff ` ` ` `return` `findMissingUtil(arr, ` `0` `, n - ` `1` `, diff); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `int` `arr[] = {` `2` `, ` `4` `, ` `8` `, ` `10` `, ` `12` `, ` `14` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println(` `"The missing element is "` `+ ` ` ` `findMissing(arr, n)); ` `} ` `} ` ` ` `// This code is contributed by anuj_67. ` |

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## C#

`// A C# program to find ` `// the missing number in ` `// a given arithmetic ` `// progression ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// A binary search based ` `// recursive function that ` `// returns the missing ` `// element in arithmetic ` `// progression ` `static` `int` `findMissingUtil(` `int` `[]arr, ` `int` `low, ` ` ` `int` `high, ` `int` `diff) ` `{ ` ` ` `// There must be two elements ` ` ` `// to find the missing ` ` ` `if` `(high <= low) ` ` ` `return` `int` `.MaxValue; ` ` ` ` ` `// Find index of ` ` ` `// middle element ` ` ` `int` `mid = low + (high - ` ` ` `low) / 2; ` ` ` ` ` `// The element just after the ` ` ` `// middle element is missing. ` ` ` `// The arr[mid+1] must exist, ` ` ` `// because we return when ` ` ` `// (low == high) and take ` ` ` `// floor of (high-low)/2 ` ` ` `if` `(arr[mid + 1] - ` ` ` `arr[mid] != diff) ` ` ` `return` `(arr[mid] + diff); ` ` ` ` ` `// The element just ` ` ` `// before mid is missing ` ` ` `if` `(mid > 0 && arr[mid] - ` ` ` `arr[mid - 1] != diff) ` ` ` `return` `(arr[mid - 1] + diff); ` ` ` ` ` `// If the elements till mid follow ` ` ` `// AP, then recur for right half ` ` ` `if` `(arr[mid] == arr[0] + ` ` ` `mid * diff) ` ` ` `return` `findMissingUtil(arr, mid + 1, ` ` ` `high, diff); ` ` ` ` ` `// Else recur for left half ` ` ` `return` `findMissingUtil(arr, low, ` ` ` `mid - 1, diff); ` `} ` ` ` `// The function uses findMissingUtil() ` `// to find the missing element ` `// in AP. It assumes that there ` `// is exactly one missing element ` `// and may give incorrect result ` `// when there is no missing element ` `// or more than one missing elements. ` `// This function also assumes that ` `// the difference in AP is an integer. ` `static` `int` `findMissing(` `int` `[]arr, ` `int` `n) ` `{ ` ` ` `// If exactly one element ` ` ` `// is missing, then we can ` ` ` `// find difference of arithmetic ` ` ` `// progression using following ` ` ` `// formula. Example, 2, 4, 6, 10, ` ` ` `// diff = (10-2)/4 = 2.The assumption ` ` ` `// in formula is that the difference ` ` ` `// is an integer. ` ` ` `int` `diff = (arr[n - 1] - ` ` ` `arr[0]) / n; ` ` ` ` ` `// Binary search for the ` ` ` `// missing number using ` ` ` `// above calculated diff ` ` ` `return` `findMissingUtil(arr, 0, ` ` ` `n - 1, diff); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `Main () ` `{ ` ` ` `int` `[]arr = {2, 4, 8, ` ` ` `10, 12, 14}; ` ` ` `int` `n = arr.Length; ` ` ` `Console.WriteLine(` `"The missing element is "` `+ ` ` ` `findMissing(arr, n)); ` `} ` `} ` ` ` `// This code is contributed by anuj_67. ` |

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## PHP

0 && $arr[$mid] – $arr[$mid – 1] != $diff)

return ($arr[$mid – 1] + $diff);

// If the elements till mid follow AP,

// then recur for right half

if ($arr[$mid] == $arr[0] + $mid * $diff)

return findMissingUtil($arr, $mid + 1,

$high, $diff);

// Else recur for left half

return findMissingUtil($arr, $low, $mid – 1, $diff);

}

// The function uses findMissingUtil() to find

// the missing element in AP. It assumes that

// there is exactly one missing element and may

// give incorrect result when there is no missing

// element or more than one missing elements.

// This function also assumes that the difference

// in AP is an integer.

function findMissing($arr, $n)

{

// If exactly one element is missing, then

// we can find difference of arithmetic

// progression using following formula.

// Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.

// The assumption in formula is that the

// difference is an integer.

$diff = ($arr[$n – 1] – $arr[0]) / $n;

// Binary search for the missing number

// using above calculated diff

return findMissingUtil($arr, 0, $n – 1, $diff);

}

// Driver Code

$arr = array(2, 4, 8, 10, 12, 14);

$n = sizeof($arr);

echo “The missing element is “,

findMissing($arr, $n);

// This code is contributed by Sach_Code

?>

**Output: **

The missing element is 6

**Exercise:**

Solve the same problem for Geometrical Series. What is the time complexity of your solution? What about Fibonacci Series?

This article is contributed by **Harshit Agrawal**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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