View Discussion Improve Article Save Article View Discussion Improve Article Save Article Given an unsorted array and an element x, search x in the given array. Write recursive C code for this. If the element is not present, return -1. Approach : The idea is to compare x with the last element in arr[].
If the element is found at the last position, return it. Else recur searchElement[] for remaining array and element x. C++
#include using namespace std;int searchElement[int arr[], int size, int x] { size--; if [size < 0] { return -1;
}
if [arr[size] == x] {
return size;
}
return searchElement[arr, size, x];
}
int main[] {
int arr[] = {17, 15, 11, 8, 13, 19};
int size = sizeof[arr] / sizeof[arr[0]];
int x = 11;
int idx = searchElement[arr, size, x];
if [idx != -1]
cout
Javascript
function recSearch[arr, l, r, x]
{
if [r < l]
return -1;
if [arr[l] == x]
return l;
if [arr[r] == x]
return r;
return recSearch[arr, l+1, r-1, x];
}
let arr = [12, 34, 54, 2, 3];
let i;
let n = arr.length;
let x = 23;
let index = recSearch[arr, 0, n - 1, x];
if [index != -1]{
document.write[`Element ${x} is present at index ${index}`];
}
else{
document.write["Element is not present " + x];
}
Output
Element 11 is present at index 2
Explanation
We iterate through the array from the end by decrementing the size variable and recursively calling the function searchElement[]. If the size variable becomes less than zero it means the element is not present in the array and we return -1. If a match is found, we return the size variable which is the index of the found element. This process is repeated until a value is returned to main[].
It is important to note that if there are duplicate elements in the array, the index of the last matched element will be returned since we are [recursively] iterating the array from the end.
Time Complexity: O[N], where N is the size of the given array.
Auxiliary Space:
O[1]
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