dMin-Max Range Queries - GeeksforGeeks
Given an array arr[0 . . . n-1]. We need to efficiently find the minimum and maximum value from index qs (query start) to qe (query end) where 0 <= qs <= qe <= n-1. We are given multiple queries.
Given an array arr[0 . . . n-1]. We need to efficiently find the minimum and maximum value from index qs (query start) to qe (query end) where 0 <= qs <= qe <= n-1. We are given multiple queries.
Efficient solution : This problem can be solved more efficiently by using Segment Tree.
Can we do better if there are no updates on array?
The above segment tree based solution also allows array updates also to happen in O(Log n) time. Assume a situation when there are no updates (or array is static). We can actually process all queries in O(1) time with some preprocessing. One simple solution is to make a 2D table of nodes that stores all range minimum and maximum. This solution requires O(1) query time, but requires O(n2) preprocessing time and O(n2) extra space which can be a problem for large n. We can solve this problem in O(1) query time, O(n Log n) space and O(n Log n) preprocessing time using Sparse Table.
The above segment tree based solution also allows array updates also to happen in O(Log n) time. Assume a situation when there are no updates (or array is static). We can actually process all queries in O(1) time with some preprocessing. One simple solution is to make a 2D table of nodes that stores all range minimum and maximum. This solution requires O(1) query time, but requires O(n2) preprocessing time and O(n2) extra space which can be a problem for large n. We can solve this problem in O(1) query time, O(n Log n) space and O(n Log n) preprocessing time using Sparse Table.
// Node for storing minimum nd maximum value of given rangestruct node{ int minimum; int maximum;};// A utility function to get the middle index from corner indexes.int getMid(int s, int e) { return s + (e -s)/2; }/* A recursive function to get the minimum and maximum value in a given range of array indexes. The following are parameters for this function. st --> Pointer to segment tree index --> Index of current node in the segment tree. Initially 0 is passed as root is always at index 0 ss & se --> Starting and ending indexes of the segment represented by current node, i.e., st[index] qs & qe --> Starting and ending indexes of query range */struct node MaxMinUntill(struct node *st, int ss, int se, int qs, int qe, int index){ // If segment of this node is a part of given range, then return // the minimum and maximum node of the segment struct node tmp,left,right; if (qs <= ss && qe >= se) return st[index]; // If segment of this node is outside the given range if (se < qs || ss > qe) { tmp.minimum = INT_MAX; tmp.maximum = INT_MIN; return tmp; } // If a part of this segment overlaps with the given range int mid = getMid(ss, se); left = MaxMinUntill(st, ss, mid, qs, qe, 2*index+1); right = MaxMinUntill(st, mid+1, se, qs, qe, 2*index+2); tmp.minimum = min(left.minimum, right.minimum); tmp.maximum = max(left.maximum, right.maximum); return tmp;}// Return minimum and maximum of elements in range from index// qs (quey start) to qe (query end). It mainly uses// MaxMinUtill()struct node MaxMin(struct node *st, int n, int qs, int qe){ struct node tmp; // Check for erroneous input values if (qs < 0 || qe > n-1 || qs > qe) { printf("Invalid Input"); tmp.minimum = INT_MIN; tmp.minimum = INT_MAX; return tmp; } return MaxMinUntill(st, 0, n-1, qs, qe, 0);}// A recursive function that constructs Segment Tree for array[ss..se].// si is index of current node in segment tree stvoid constructSTUtil(int arr[], int ss, int se, struct node *st, int si){ // If there is one element in array, store it in current node of // segment tree and return if (ss == se) { st[si].minimum = arr[ss]; st[si].maximum = arr[ss]; return ; } // If there are more than one elements, then recur for left and // right subtrees and store the minimum and maximum of two values // in this node int mid = getMid(ss, se); constructSTUtil(arr, ss, mid, st, si*2+1); constructSTUtil(arr, mid+1, se, st, si*2+2); st[si].minimum = min(st[si*2+1].minimum, st[si*2+2].minimum); st[si].maximum = max(st[si*2+1].maximum, st[si*2+2].maximum);}/* Function to construct segment tree from given array. This function allocates memory for segment tree and calls constructSTUtil() to fill the allocated memory */struct node *constructST(int arr[], int n){ // Allocate memory for segment tree // Height of segment tree int x = (int)(ceil(log2(n))); // Maximum size of segment tree int max_size = 2*(int)pow(2, x) - 1; struct node *st = new struct node[max_size]; // Fill the allocated memory st constructSTUtil(arr, 0, n-1, st, 0); // Return the constructed segment tree return st;}
We have an array arr[0 . . . n-1]. We should be able to efficiently find the minimum value from index L (query start) to R (query end) where 0 <= L <= R <= n-1. Consider a situation when there are many range queries.
Method 2 (Square Root Decomposition)
We can use Square Root Decompositions to reduce space required in above method.
We can use Square Root Decompositions to reduce space required in above method.
Preprocessing:
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute minimum of every block of size √n and store the results.
1) Divide the range [0, n-1] into different blocks of √n each.
2) Compute minimum of every block of size √n and store the results.
Preprocessing takes O(√n * √n) = O(n) time and O(√n) space.
Query:
1) To query a range [L, R], we take minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with given range, we linearly scan them to find minimum.
1) To query a range [L, R], we take minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with given range, we linearly scan them to find minimum.
Time complexity of query is O(√n). Note that we have minimum of middle block directly accessible and there can be at most O(√n) middle blocks. There can be atmost two corner blocks that we may have to scan, so we may have to scan 2*O(√n) elements of corner blocks. Therefore, overall time complexity is O(√n).
Method 3 (Sparse Table Algorithm)
The above solution requires only O(√n) space, but takes O(√n) time to query. Sparse table method supports query time O(1) with extra space O(n Log n).
The above solution requires only O(√n) space, but takes O(√n) time to query. Sparse table method supports query time O(1) with extra space O(n Log n).
The idea is to precompute minimum of all subarrays of size 2j where j varies from 0 to Log n. Like method 1, we make a lookup table. Here lookup[i][j] contains minimum of range starting from i and of size 2j. For example lookup[0][3] contains minimum of range [0, 7] (starting with 0 and of size 23)
Preprocessing:
How to fill this lookup table? The idea is simple, fill in bottom up manner using previously computed values.
How to fill this lookup table? The idea is simple, fill in bottom up manner using previously computed values.
For example, to find minimum of range [0, 7], we can use minimum of following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]
Based on above example, below is formula,
// If arr[lookup[0][3]] <= arr[lookup[4][7]], // then lookup[0][7] = lookup[0][3] If arr[lookup[i][j-1]] <= arr[lookup[i+2j-1-1][j-1]] lookup[i][j] = lookup[i][j-1] // If arr[lookup[0][3]] > arr[lookup[4][7]], // then lookup[0][7] = lookup[4][7] Else lookup[i][j] = lookup[i+2j-1-1][j-1]
Query:
For any arbitrary range [l, R], we need to use ranges which are in powers of 2. The idea is to use closest power of 2. We always need to do at most comparison (compare minimum of two ranges which are powers of 2). One range starts with L and and ends with “L + closest-power-of-2″. The other range ends at R and starts with “R – same-closest-power-of-2 + 1″. For example, if given range is (2, 10), we compare minimum of two ranges (2, 9) and (3, 10).
For any arbitrary range [l, R], we need to use ranges which are in powers of 2. The idea is to use closest power of 2. We always need to do at most comparison (compare minimum of two ranges which are powers of 2). One range starts with L and and ends with “L + closest-power-of-2″. The other range ends at R and starts with “R – same-closest-power-of-2 + 1″. For example, if given range is (2, 10), we compare minimum of two ranges (2, 9) and (3, 10).
Based on above example, below is formula,
// For (2,10), j = floor(Log2(10-2+1)) = 3 j = floor(Log(R-L+1)) // If arr[lookup[0][7]] <= arr[lookup[3][10]], // then RMQ(2,10) = lookup[0][7] If arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]] RMQ(L, R) = lookup[L][j] // If arr[lookup[0][7]] > arr[lookup[3][10]], // then RMQ(2,10) = lookup[3][10] Else RMQ(L, R) = lookup[i+2j-1-1][j-1]
Since we do only one comparison, time complexity of query is O(1).
So sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time and O(n Log n) space.
// lookup[i][j] is going to store index of minimum value in// arr[i..j]. Ideally lookup table size should not be fixed and// should be determined using n Log n. It is kept constant to// keep code simple.int lookup[MAX][MAX];// Structure to represent a query rangestruct Query{ int L, R;};// Fills lookup array lookup[][] in bottom up manner.void preprocess(int arr[], int n){ // Initialize M for the intervals with length 1 for (int i = 0; i < n; i++) lookup[i][0] = i; // Compute values from smaller to bigger intervals for (int j=1; (1<<j)<=n; j++) { // Compute minimum value for all intervals with size 2^j for (int i=0; (i+(1<<j)-1) < n; i++) { // For arr[2][10], we compare arr[lookup[0][7]] and // arr[lookup[3][10]] if (arr[lookup[i][j-1]] < arr[lookup[i + (1<<(j-1))][j-1]]) lookup[i][j] = lookup[i][j-1]; else lookup[i][j] = lookup[i + (1 << (j-1))][j-1]; } }}// Returns minimum of arr[L..R]int query(int arr[], int L, int R){ // For [2,10], j = 3 int j = (int)log2(R-L+1); // For [2,10], we compare arr[lookup[0][7]] and // arr[lookup[3][10]], if (arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]]) return arr[lookup[L][j]]; else return arr[lookup[R-(int)pow(2,j)+1][j]];}// Prints minimum of given m query ranges in arr[0..n-1]void RMQ(int arr[], int n, Query q[], int m){ // Fills table lookup[n][Log n] preprocess(arr, n); // One by one compute sum of all queries for (int i=0; i<m; i++) { // Left and right boundaries of current range int L = q[i].L, R = q[i].R; // Print sum of current query range cout << "Minimum of [" << L << ", " << R << "] is " << query(arr, L, R) << endl; }}
This approach supports query in O(1), but preprocessing takes O(n2) time. Also, this approach needs O(n2) extra space which may become huge for large input arrays.
// lookup[i][j] is going to store index of minimum value in// arr[i..j]int lookup[MAX][MAX];// Structure to represent a query rangestruct Query{ int L, R;};// Fills lookup array lookup[n][n] for all possible values of// query rangesvoid preprocess(int arr[], int n){ // Initialize lookup[][] for the intervals with length 1 for (int i = 0; i < n; i++) lookup[i][i] = i; // Fill rest of the entries in bottom up manner for (int i=0; i<n; i++) { for (int j = i+1; j<n; j++) // To find minimum of [0,4], we compare minimum of // arr[lookup[0][3]] with arr[4]. if (arr[lookup[i][j - 1]] < arr[j]) lookup[i][j] = lookup[i][j - 1]; else lookup[i][j] = j; }}// Prints minimum of given m query ranges in arr[0..n-1]void RMQ(int arr[], int n, Query q[], int m){ // Fill lookup table for all possible input queries preprocess(arr, n); // One by one compute sum of all queries for (int i=0; i<m; i++) { // Left and right boundaries of current range int L = q[i].L, R = q[i].R; // Print sum of current query range cout << "Minimum of [" << L << ", " << R << "] is " << arr[lookup[L][R]] << endl; }}
Simple Solution : We solve this problem using Tournament Method for each query. Complexity for this approach will be O(queries * n).
http://www.geeksforgeeks.org/maximum-and-minimum-in-an-array/
Maximum and minimum of an array using minimum number of comparisons
Write a C function to return minimum and maximum in an array. You program should make minimum number of comparisons.
METHOD 2 (Tournament Method)
Divide the array into two parts and compare the maximums and minimums of the the two parts to get the maximum and the minimum of the the whole array.
Divide the array into two parts and compare the maximums and minimums of the the two parts to get the maximum and the minimum of the the whole array.
Pair MaxMin(array, array_size)
if array_size = 1
return element as both max and min
else if arry_size = 2
one comparison to determine max and min
return that pair
else /* array_size > 2 */
recur for max and min of left half
recur for max and min of right half
one comparison determines true max of the two candidates
one comparison determines true min of the two candidates
return the pair of max and min
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