# Closest greater element for every array element from another array

Given two arrays a[] and b[], we need to build an array c[] such that every element c[i] of c[] contains a value from a[] which is greater than b[i] and is closest to b[i]. If a[] has no greater element than b[i], then value of c[i] is -1. All arrays are of same size.

Examples:

Input : a[] = [ 2, 6, 5, 7, 0] b[] = [1, 3, 2, 5, 8] Output : c[] = [2, 5, 5, 7, -1] Input : a[] = [ 2, 6, 5, 7, 0] b[] = [0, 2, 3, 5, 1] Output : c[] = [2, 5, 5, 6, 2]

**Naive Approach :** For each element in b[], we traverse the whole of a[] and try to find the closest greater element, and save the result for each search. This will cost time complexity of O(n^2).

**Efficient Approach :** Sort the array a[], and for each b[i], apply binary search in sorted array a[]. For this method our time complexity will be O(nlogn).

**Note:** For closest greater element we can use upper_bound().

`// CPP to find result from target array ` `// for closest element ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function for printing resultant array ` `void` `closestResult(` `int` `a[], ` `int` `b[], ` `int` `n) ` `{ ` ` ` `// change arr[] to vector ` ` ` `vector<` `int` `> vect(a, a + n); ` ` ` ` ` `// sort vector for ease ` ` ` `sort(vect.begin(), vect.end()); ` ` ` ` ` `// iterator for upper_bound ` ` ` `vector<` `int` `>::iterator up; ` ` ` ` ` `// vector for result ` ` ` `vector<` `int` `> c; ` ` ` ` ` `// calculate resultant array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// check upper bound element ` ` ` `up = upper_bound(vect.begin(), vect.end(), b[i]); ` ` ` ` ` `// if no element found push -1 ` ` ` `if` `(up == vect.end()) ` ` ` `c.push_back(-1); ` ` ` ` ` `// Else push the element ` ` ` `else` ` ` `c.push_back(*up); ` `// add to resultant ` ` ` `} ` ` ` ` ` `cout << ` `"Result = "` `; ` ` ` `for` `(` `auto` `it = c.begin(); it != c.end(); it++) ` ` ` `cout << *it << ` `" "` `; ` `} ` ` ` `// driver program ` `int` `main() ` `{ ` ` ` `int` `a[] = { 2, 5, 6, 1, 8, 9 }; ` ` ` `int` `b[] = { 2, 1, 0, 5, 4, 9 }; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `closestResult(a, b, n); ` ` ` `return` `0; ` `} ` |

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Output:

Result = 5 2 1 6 5 -1

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