Given an array of n elements and q range queries (range sum in this article) with no updates, task is to answer these queries with efficient time and space complexity. The time complexity of a range query after applying square root decomposition comes out to be **O(√n)**. This square-root factor can be decreased to a constant linear factor by applying square root decomposition on the block of the array which was decomposed earlier.

**Prerequisite: ** Mo’s Algorithm | Prefix Array

**Approach :**

As we apply square root decomposition to the given array, querying a range-sum comes in **O(√n)** time.

Here, calculate the sum of blocks which are in between the blocks under consideration(corner blocks), which takes **O(√n)** iterations.

Initial Array :

Decomposition of array into blocks :

And the calculation time for the sum on the starting block and ending block both takes **O(√n)** iterations.

**Which will leaves us per query time complexity of :**

= O(√n) + O(√n) + O(√n) = 3 * O(√n) ~O(√n)

Here, we can reduce the runtime complexity of our query algorithm cleverly by calculating blockwise prefix sum and using it to calculate the sum accumulated in the blocks which lie between the blocks under consideration. Consider the code below :

interblock_sum[x1][x2] = prefixData[x2 - 1] - prefixData[x1];

**Time taken for calculation of above table is : **

= O(√n) * O(√n) ~O(n)

**NOTE :** We haven’t taken the sum of blocks x1 & x2 under consideration as they might be carrying partial data.

**Prefix Array :**

Suppose we want to query for the sum for range from 4 to 11, we consider the sum between block 0 and block 2 (excluding the data contained in block 0 and block 1), which can be calculated using the sum in the green coloured blocks represented in the above image.

Sum between block 0 and block 2 = 42 – 12 = 30

For calculation of rest of the sum present in the yellow blocks, consider the prefix array at the decomposition level-2 and repeat the process again.

Here, observe that we have reduced our time complexity per query significantly, though our runtime remains similar to our last approach :

**Our new time complexity can be calculated as :**

= O(√n) + O(1) + O(√n) = 2 * O(√n) ~O(√n)

Square-root Decomposition at Level-2 :

Further, we apply square root decomposition again on every decomposed block retained from the previous decomposition. Now at this level, we have approximately **√√n** sub-blocks in each block which were decomposed at last level. So, we need to run a range query on these blocks only two times, one time for starting block and one time for ending block.

**Precalcuation Time taken for level 2 decomposition :**

No of blocks at level 1 ~ **√n**

No of blocks at level 2 ~ **√√n**

**Level-2 Decomposition Runtime of a level-1 decomposed block :**

= O(√n)

**Overall runtime of level-2 decomposition over all blocks :**

= O(√n) * O(√n) ~O(n)

Now, we can query our level-2 decomposed blocks in **O(√√n)** time.

So, we have reduced our overall time complexity from **O(√n)** to **O(√√n)**

**Time complxity taken in querying edge blocks :**

= O(√√n) + O(1) + O(√√n) = 2 * O(√√n) ~O(√√n)

**Total Time complexity can be calculated as :**

= O(√√n)+O(1)+O(√√n) = 2 * O(√√n) ~O(√√n)

Square-root Decomposition at Level-3 :

Using this method we can decompose our array again and again recursively **d times** to reduce our time complexity to a factor of **constant linearity**.

O(d * n^{1/(2^d)}) ~ O(k), as d increases this factor converges to a constant linear term

The code presented below is a representation of triple square root decomposition where d = 3:

O(q * d * n^{1/(2 ^ 3)}) ~ O(q * k) ~ O(q)

[where q repesents number of range queries]

// CPP code for offline queries in // approx constant time. #include<bits/stdc++.h> using namespace std; int n1; // Structure to store decomposed data typedef struct { vector<int> data; vector<vector<int>> rdata; int blocks; int blk_sz; }sqrtD; vector<vector<sqrtD>> Sq3; vector<sqrtD> Sq2; sqrtD Sq1; // Square root Decomposition of // a given array sqrtD decompose(vector<int> arr) { sqrtD sq; int n = arr.size(); int blk_idx = -1; sq.blk_sz = sqrt(n); sq.data.resize((n/sq.blk_sz) + 1, 0); // Calculation of data in blocks for (int i = 0; i < n; i++) { if (i % sq.blk_sz == 0) { blk_idx++; } sq.data[blk_idx] += arr[i]; } int blocks = blk_idx + 1; sq.blocks = blocks; // Calculation of prefix data int prefixData[blocks]; prefixData[0] = sq.data[0]; for(int i = 1; i < blocks; i++) { prefixData[i] = prefixData[i - 1] + sq.data[i]; } sq.rdata.resize(blocks + 1, vector<int>(blocks + 1)); // Calculation of data between blocks for(int i = 0 ;i < blocks; i++) { for(int j = i + 1; j < blocks; j++) { sq.rdata[i][j] = sq.rdata[j][i] = prefixData[j - 1] - prefixData[i]; } } return sq; } // Sqaure root Decompostion at level3 vector<vector<sqrtD>> tripleDecompose(sqrtD sq1, sqrtD sq2,vector<int> &arr) { vector<vector<sqrtD>> sq(sq1.blocks, vector<sqrtD>(sq1.blocks)); int blk_idx1 = -1; for(int i = 0; i < sq1.blocks; i++) { int blk_ldx1 = blk_idx1 + 1; blk_idx1 = (i + 1) * sq1.blk_sz - 1; blk_idx1 = min(blk_idx1,n1 - 1); int blk_idx2 = blk_ldx1 - 1; for(int j = 0; j < sq2.blocks; ++j) { int blk_ldx2 = blk_idx2 + 1; blk_idx2 = blk_ldx1 + (j + 1) * sq2.blk_sz - 1; blk_idx2 = min(blk_idx2, blk_idx1); vector<int> ::iterator it1 = arr.begin() + blk_ldx2; vector<int> ::iterator it2 = arr.begin() + blk_idx2 + 1; vector<int> vec(it1, it2); sq[i][j] = decompose(vec); } } return sq; } // Sqaure root Decompostion at level2 vector<sqrtD> doubleDecompose(sqrtD sq1, vector<int> &arr) { vector<sqrtD> sq(sq1.blocks); int blk_idx = -1; for(int i = 0; i < sq1.blocks; i++) { int blk_ldx = blk_idx + 1; blk_idx = (i + 1) * sq1.blk_sz - 1; blk_idx = min(blk_idx, n1 - 1); vector<int> ::iterator it1 = arr.begin() + blk_ldx; vector<int> ::iterator it2 = arr.begin() + blk_idx + 1; vector<int> vec(it1, it2); sq[i] = decompose(vec); } return sq; } // Sqaure root Decompostion at level1 void singleDecompose(vector<int> &arr) { sqrtD sq1 = decompose(arr); vector<sqrtD> sq2(sq1.blocks); sq2 = doubleDecompose(sq1, arr); vector<vector<sqrtD>> sq3(sq1.blocks, vector<sqrtD>(sq2[0].blocks)); sq3 = tripleDecompose(sq1, sq2[0],arr); // ASSIGNMENT TO GLOBAL VARIABLES Sq1 = sq1; Sq2.resize(sq1.blocks); Sq2 = sq2; Sq3.resize(sq1.blocks, vector<sqrtD>(sq2[0].blocks)); Sq3 = sq3; } // Function for query at level 3 int queryLevel3(int start,int end, int main_blk, int sub_main_blk, vector<int> &arr) { int blk_sz= Sq3[0][0].blk_sz; // Element Indexing at level2 decompostion int nstart = start - main_blk * Sq1.blk_sz - sub_main_blk * Sq2[0].blk_sz; int nend = end - main_blk * Sq1.blk_sz - sub_main_blk * Sq2[0].blk_sz; // Block indexing at level3 decompostion int st_blk = nstart / blk_sz; int en_blk = nend / blk_sz; int answer = Sq3[main_blk][sub_main_blk].rdata[st_blk][en_blk]; // If start and end point dont lie in same block if(st_blk != en_blk) { int left = 0, en_idx = main_blk * Sq1.blk_sz + sub_main_blk * Sq2[0].blk_sz + (st_blk + 1) * blk_sz -1; for(int i = start; i <= en_idx; i++) { left += arr[i]; } int right = 0, st_idx = main_blk * Sq1.blk_sz + sub_main_blk * Sq2[0].blk_sz + (en_blk) * blk_sz; for(int i = st_idx; i <= end; i++) { right += arr[i]; } answer += left; answer += right; } else { for(int i = start; i <= end; i++) { answer += arr[i]; } } return answer; } // Function for splitting query to level two int queryLevel2(int start, int end, int main_blk, vector<int> &arr) { int blk_sz = Sq2[0].blk_sz; // Element Indexing at level1 decompostion int nstart = start - (main_blk * Sq1.blk_sz); int nend = end - (main_blk * Sq1.blk_sz); // Block indexing at level2 decompostion int st_blk = nstart / blk_sz; int en_blk = nend / blk_sz; // Interblock data level2 decompostion int answer = Sq2[main_blk].rdata[st_blk][en_blk]; if(st_blk == en_blk) { answer += queryLevel3(start, end, main_blk, st_blk, arr); } else { answer += queryLevel3(start, (main_blk * Sq1.blk_sz) + ((st_blk + 1) * blk_sz) - 1, main_blk, st_blk, arr); answer += queryLevel3((main_blk * Sq1.blk_sz) + (en_blk * blk_sz), end, main_blk, en_blk, arr); } return answer; } // Function to return answer according to query int Query(int start,int end,vector<int>& arr) { int blk_sz = Sq1.blk_sz; int st_blk = start / blk_sz; int en_blk = end / blk_sz; // Interblock data level1 decompostion int answer = Sq1.rdata[st_blk][en_blk]; if(st_blk == en_blk) { answer += queryLevel2(start, end, st_blk, arr); } else { answer += queryLevel2(start, (st_blk + 1) * blk_sz - 1, st_blk, arr); answer += queryLevel2(en_blk * blk_sz, end, en_blk, arr); } // returning final answer return answer; } // Driver code int main() { n1 = 16; vector<int> arr = {7, 2, 3, 0, 5, 10, 3, 12, 18, 1, 2, 3, 4, 5, 6, 7}; singleDecompose(arr); int q = 5; pair<int, int> query[q] = {{6, 10}, {7, 12}, {4, 13}, {4, 11}, {12, 16}}; for(int i = 0; i < q; i++) { int a = query[i].first, b = query[i].second; printf("%d\n", Query(a - 1, b - 1, arr)); } return 0; }

**Output:**

44 39 58 51 25

**Time Complexity : **O(q * d * n^{1/(2^3)}) ≈ O(q * k) ≈ O(q)

**Auxiliary Space : **O(k * n) ≈ O(n)

**Note : **This article is to only explain the method of decomposing the square root to further decomposition.

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