Given a limited range array contains both positive and non positive numbers, i.e., elements are in range from -MAX to +MAX. Our task is to search for some number is present in the array of not in O(1) time.

Since range is limited, we can use index mapping (or trivial hashing). We use values as index in a big array. Therefore we can search and insert elements in O(1) time.

**How to handle negative numbers?**

The idea is to use a 2D array of size hash[MAX+1][2]

*Algorithm:*

Assign all the values of the hash matrix as 0. Traverse the given array: If the elementeleis non negative assign hash[ele][0] as 1. Else take the absolute value ofeleand assign hash[ele][1] as 1.

To search any element *x* in the array.

- If X is non-negative check if hash[X][0] is 1 or not. If hash[X][0] is one then the number is present else not present.
- If X is negative take absolute vale of X and then check if hash[X][1] is 1 or not. If hash[X][1] is one then the number is present

Below is the implementation of the above idea.

// CPP program to implement direct index mapping // with negative values allowed. #include <bits/stdc++.h> using namespace std; #define MAX 1000 // Since array is global, it is initialized as 0. bool has[MAX + 1][2]; // searching if X is Present in the given array // or not. bool search(int X) { if (X >= 0) { if (has[X][0] == 1) return true; else return false; } // if X is negative take the absolute // value of X. X = abs(X); if (has[X][1] == 1) return true; return false; } void insert(int a[], int n) { for (int i = 0; i < n; i++) { if (a[i] >= 0) has[a[i]][0] = 1; else has[abs(a[i])][1] = 1; } } // Driver code int main() { int a[] = { -1, 9, -5, -8, -5, -2 }; int n = sizeof(a)/sizeof(a[0]); int X = -5; if (search(X) == true) cout << "Present"; else cout << "Not Present"; return 0; }

Output:

Present

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