Given an array A[ ] consisting of non-negative integers and matrix Q[ ][ ] consisting of queries of the following two types:

• **(1, l, val): **Update A[l] to A[l] + val.
• **(2, K): **Find the lower_bound of **K **in the prefix sum array of A[ ]. If the lower_bound does not exist print -1.

The task for each query of second type is to print the index of lower_bound of value K.

Examples:

_Input: __A[ ] = {1, 2, 3, 5, 8}, Q[ ][ ] = {{1, 0, 2}, {2, 5}, {1, 3, 5}} _

_Output: __1 _

Explanation:

Query 1: Update A[0] to A[0] + 2. Now A[ ] = {3, 2, 3, 5, 8}

_Query 2: lower_bound of K = 5 in the prefix sum array {3, 5, 8, 13, 21} is 5 and index = 1. _

Query 3: Update A[3] to A[3] + 5. Now A[ ] = {3, 2, 3, 10, 8}

_Input: __A[ ] = {4, 1, 12, 8, 20}, Q[ ] = {{2, 50}, {1, 3, 12}, {2, 50}} _

_Output: __-1 _

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach:

The simplest approach is to firstly build a prefix sum array of given array A[ ], and for queries of Type 1, update values and recalculate the prefix sum. For query of Type 2, perform a Binary Search on the prefix sum array to find lower bound.

Time Complexity:_ O(Q*(N*logn))_

_Auxiliary Space: _O(N)

Efficient Approach:

The above approach can be optimized using Fenwick Tree. Using this Data Structure, the update queries in prefix sum array can be performed in logarithmic time.

Follow the steps below to solve the problem:

• Construct the Prefix Sum Array using Fenwick Tree.
• For queries of Type 1, while** l > 0**, add val to A[l] traverse to the parent node by adding least significant bit in l.
• For queries of Type 2, perform the Binary Search on the Fenwick Tree to obtain the lower bound.
• Whenever a prefix sum greater than **K appears, **store that **index **and traverse the left part of the Fenwick Tree. Otherwise, traverse the right part of the Fenwick Tree Now, perform Binary Search.
• Finally, print the required index.

Below is the implementation of the above approach:

• Java
• C#

`// Java program to implement`

`// the above appraoch`

`**import**` `java.util.*;`

`**import**` `java.io.*;`

`**class**` `GFG {`

`// Function to calculate and return`

`// the sum of arr[0..index]`

`**static**` `**int**` `getSum(``**int**` `BITree[],`

`**int**` `index)`

`{`

`**int**` `ans =` `0``;`

`index +=` `1``;`

`// Traverse ancestors`

`// of BITree[index]`

`**while**` `(index >` `0``) {`

`// Update the sum of current`

`// element of BIT to ans`

`ans += BITree[index];`

`// Update index to that`

`// of the parent node in`

`// getSum() view by`

`// subtracting LSB(Least`

`// Significant Bit)`

`index -= index & (-index);`

`}`

`**return**` `ans;`

`}`

`// Function to update the Binary Index`

`// Tree by replacing all ancestores of`

`// index by their respective sum with val`

`**static**` `**void**` `updateBIT(``**int**` `BITree[],`

`**int**` `n,` `**int**` `index,` `**int**` `val)`

`{`

`index = index +` `1``;`

`// Traverse all ancestors`

`// and sum with 'val'.`

`**while**` `(index <= n) {`

`// Add 'val' to current`

`// node of BIT`

`BITree[index] += val;`

`// Update index to that`

`// of the parent node in`

`// updateBit() view by`

`// adding LSB(Least`

`// Significant Bit)`

`index += index & (-index);`

`}`

`}`

`// Function to construct the Binary`

`// Indexed Tree for the given array`

`**static**` `**int**``[] constructBITree(`

`**int**` `arr[],` `**int**` `n)`

`{`

`// Initialize the`

`// Binary Indexed Tree`

`**int**``[] BITree =` `**new**` `**int**``[n +` `1``];`

`**for**` `(``**int**` `i =` `0``; i <= n; i++)`

`BITree[i] =` `0``;`

`// Store the actual values in`

`// BITree[] using update()`

`**for**` `(``**int**` `i =` `0``; i < n; i++)`

`updateBIT(BITree, n, i, arr[i]);`

`**return**` `BITree;`

`}`

`// Function to obtian and return`

`// the index of lower_bound of k`

`**static**` `**int**` `getLowerBound(``**int**` `BITree[],`

`**int**``[] arr,` `**int**` `n,` `**int**` `k)`

`{`

`**int**` `lb = -``1``;`

`**int**` `l =` `0``, r = n -` `1``;`

`**while**` `(l <= r) {`

`**int**` `mid = l + (r - l) /` `2``;`

`**if**` `(getSum(BITree, mid) >= k) {`

`r = mid -` `1``;`

`lb = mid;`

`}`

`**else**`

`l = mid +` `1``;`

`}`

`**return**` `lb;`

`}`

`**static**` `**void**` `performQueries(``**int**` `A[],` `**int**` `n,` `**int**` `q[][])`

`{`

`// Store the Binary Indexed Tree`

`**int**``[] BITree = constructBITree(A, n);`

`// Solve each query in Q`

`**for**` `(``**int**` `i =` `0``; i < q.length; i++) {`

`**int**` `id = q[i][``0``];`

`**if**` `(id ==` `1``) {`

`**int**` `idx = q[i][``1``];`

`**int**` `val = q[i][``2``];`

`A[idx] += val;`

`// Update the values of all`

`// ancestors of idx`

`updateBIT(BITree, n, idx, val);`

`}`

`**else**` `{`

`**int**` `k = q[i][``1``];`

`**int**` `lb = getLowerBound(`

`BITree, A, n, k);`

`System.out.println(lb);`

`}`

`}`

`}`

`// Driver Code`

`**public**` `**static**` `**void**` `main(String[] args)`

`{`

`**int**` `A[] = {` `1``,` `2``,` `3``,` `5``,` `8` `};`

`**int**` `n = A.length;`

`**int**``[][] q = { {` `1``,` `0``,` `2` `},`

`{` `2``,` `5` `},`

`{` `1``,` `3``,` `5` `} };`

`performQueries(A, n, q);`

`}`

`}`

Output:

``````1
``````

Time Complexity:_ O(Q*(logN)2)_

Auxiliary Space:_ O(N)_

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

#advanced data structure #arrays #bit magic #mathematical #searching #array-range-queries #bit #prefix-sum

2.30 GEEK