### PROBLEM LINK:

**Author:** Aniket Marlapalle

**Tester:** Harsh Shah

**Editorialist:** Aniket Marlapalle

### DIFFICULTY:

easy

### PREREQUISITES:

binary indexed trees, binary search

### PROBLEM:

You have an array of n numbers. You need to perform q queries of following type:

- Change the value at
**x**index to^{th}**Y** - Find if there is a suffix with sum equal to some given value
**Y**

### EXPLANATION:

Brute-force : For every query of the first type just update the value in the array. For every second query iterate from the end of the array to see if the suffix sum is equal to some given value.

Now we can see that the suffix sums if ordered from right to left follow the non-decreasing order. So we can use **binary search** to speed up the calculations. Do a binary search on the length of the suffix and check the sum of the suffix corresponding to this length, if it is greater that the required sum then check on lower lengths else continue the search on higher lengths.

Now the only thing required is to find and update all the prefix sums. For this we can consider the reverse array and use **Binary indexed trees** or **segment trees** to update the array values and find the suffix sum faster.

**Time Complexity** : O(N*log(N) + Q*log(N)*log(N))

First creation of BIT takes O(log(N)) time for each element. For every query binary search takes O(log(N)) steps and every BIT query is O(log(N)).

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here