Range Minimum Query (Square Root Decomposition and Sparse Table)
Last Updated :
23 Jul, 2025
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.
Example:
Input: arr[] = {7, 2, 3, 0, 5, 10, 3, 12, 18};
query[] = [0, 4], [4, 7], [7, 8]Output: Minimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12
A simple solution is to run a loop from L to R and find the minimum element in the given range. This solution takes O(n) time to query in the worst case.
Another approach is to use Segment tree. With segment tree, preprocessing time is O(n) and time to for range minimum query is O(Logn). The extra space required is O(n) to store the segment tree. Segment tree allows updates also in O(Log n) time.
Can we do better if we know that the array is static?
How to optimize query time when there are no update operations and there are many range minimum queries?
Below are different methods.
Method 1 (Simple Solution)
A Simple Solution is to create a 2D array lookup[][] where an entry lookup[i][j] stores the minimum value in range arr[i..j]. The minimum of a given range can now be calculated in O(1) time.
C++
// C++ program to do range
// minimum query in O(1) time with
// O(n*n) extra space and O(n*n)
// preprocessing time.
#include <bits/stdc++.h>
using namespace std;
#define MAX 500
// lookup[i][j] is going to store
// index of minimum value in
// arr[i..j]
int lookup[MAX][MAX];
// Structure to represent a query range
struct Query {
int L, R;
};
// Fills lookup array lookup[n][n]
// for all possible values
// of query ranges
void 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;
}
}
// Driver code
int main()
{
int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = sizeof(a) / sizeof(a[0]);
Query q[] = { { 0, 4 }, { 4, 7 }, { 7, 8 } };
int m = sizeof(q) / sizeof(q[0]);
RMQ(a, n, q, m);
return 0;
}
// Java program to do range minimum query
// in O(1) time with O(n*n) extra space
// and O(n*n) preprocessing time.
import java.util.*;
class GFG {
static int MAX = 500;
// lookup[i][j] is going to store index of
// minimum value in arr[i..j]
static int[][] lookup = new int[MAX][MAX];
// Structure to represent a query range
static class Query {
int L, R;
public Query(int L, int R)
{
this.L = L;
this.R = R;
}
};
// Fills lookup array lookup[n][n] for
// all possible values of query ranges
static void 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]
static 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
System.out.println("Minimum of [" + L + ", " + R
+ "] is "
+ arr[lookup[L][R]]);
}
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = a.length;
Query q[] = { new Query(0, 4), new Query(4, 7),
new Query(7, 8) };
int m = q.length;
RMQ(a, n, q, m);
}
}
// This code is contributed by 29AjayKumar
# Python3 program to do range
# minimum query in O(1) time with
# O(n*n) extra space and O(n*n)
# preprocessing time.
MAX = 500
# lookup[i][j] is going to store
# index of minimum value in
# arr[i..j]
lookup = [[0 for j in range(MAX)]
for i in range(MAX)]
# Structure to represent
# a query range
class Query:
def __init__(self, L, R):
self.L = L
self.R = R
# Fills lookup array lookup[n][n]
# for all possible values
# of query ranges
def preprocess(arr, n):
# Initialize lookup[][] for the
# intervals with length 1
for i in range(n):
lookup[i][i] = i;
# Fill rest of the entries in
# bottom up manner
for i in range(n):
for j in range(i + 1, n):
# 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]
def RMQ(arr, n, q, m):
# Fill lookup table for
# all possible input queries
preprocess(arr, n);
# One by one compute sum of
# all queries
for i in range(m):
# Left and right boundaries
# of current range
L = q[i].L
R = q[i].R;
# Print sum of current query range
print("Minimum of [" + str(L) + ", " +
str(R) + "] is " +
str(arr[lookup[L][R]]))
# Driver code
if __name__ == "__main__":
a = [7, 2, 3, 0, 5,
10, 3, 12, 18]
n = len(a)
q = [Query(0, 4),
Query(4, 7),
Query(7, 8)]
m = len(q)
RMQ(a, n, q, m);
# This code is contributed by Rutvik_56
// C# program to do range minimum query
// in O(1) time with O(n*n) extra space
// and O(n*n) preprocessing time.
using System;
class GFG {
static int MAX = 500;
// lookup[i][j] is going to store index of
// minimum value in arr[i..j]
static int[, ] lookup = new int[MAX, MAX];
// Structure to represent a query range
public class Query {
public int L, R;
public Query(int L, int R)
{
this.L = L;
this.R = R;
}
};
// Fills lookup array lookup[n][n] for
// all possible values of query ranges
static void 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]
static 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
Console.WriteLine("Minimum of [" + L + ", " + R
+ "] is "
+ arr[lookup[L, R]]);
}
}
// Driver Code
public static void Main(String[] args)
{
int[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = a.Length;
Query[] q = { new Query(0, 4), new Query(4, 7),
new Query(7, 8) };
int m = q.Length;
RMQ(a, n, q, m);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to do range minimum query
// in O(1) time with O(n*n) extra space
// and O(n*n) preprocessing time
let MAX = 500;
// lookup[i][j] is going to store index of
// minimum value in arr[i..j]
let lookup = new Array(MAX);
for(let i = 0; i < MAX; i++)
{
lookup[i] = new Array(MAX);
for(let j = 0; j < MAX; j++)
lookup[i][j] = 0;
}
// Structure to represent a query range
class Query
{
constructor(L, R)
{
this.L = L;
this.R = R;
}
}
// Fills lookup array lookup[n][n] for
// all possible values of query ranges
function preprocess(arr, n)
{
// Initialize lookup[][] for
// the intervals with length 1
for (let i = 0; i < n; i++)
lookup[i][i] = i;
// Fill rest of the entries in bottom up manner
for (let i = 0; i < n; i++) {
for (let 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]
function RMQ(arr,n,q,m)
{
// Fill lookup table for
// all possible input queries
preprocess(arr, n);
// One by one compute sum of all queries
for (let i = 0; i < m; i++)
{
// Left and right boundaries
// of current range
let L = q[i].L, R = q[i].R;
// Print sum of current query range
document.write("Minimum of [" + L + ", " + R
+ "] is "
+ arr[lookup[L][R]]+"<br>");
}
}
// Driver Code
let a=[7, 2, 3, 0, 5, 10, 3, 12, 18];
let n = a.length;
let q = [ new Query(0, 4), new Query(4, 7),
new Query(7, 8) ];
let m = q.length;
RMQ(a, n, q, m);
// This code is contributed by avanitrachhadiya2155
</script>
Output:
Minimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12
This approach supports queries 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.
Method 2 (Square Root Decomposition)
We can use Square Root Decompositions to reduce space required in the above method.
Preprocessing:
1) Divide the range [0, n-1] into different blocks of ?n each.
2) Compute the 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 a minimum of all blocks that lie in this range. For left and right corner blocks which may partially overlap with the given range, we linearly scan them to find the minimum.
The time complexity of the query is O(?n). Note that we have a minimum of the middle block directly accessible and there can be at most O(?n) middle blocks. There can be at most two corner blocks that we may have to scan, so we may have to scan 2*O(?n) elements of corner blocks. Therefore, the overall time complexity is O(?n).
Refer to Sqrt (or Square Root) Decomposition Technique | Set 1 (Introduction) for details.
Method 3 (Sparse Table Algorithm)
The above solution requires only O(?n) space but takes O(?n) time to query. The sparse table method supports query time O(1) with extra space O(n Log n).
The idea is to precompute a 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 a minimum of range starting from i and of size 2j. For example lookup[0][3] contains a 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 a bottom-up manner using previously computed values.
For example, to find a minimum of range [0, 7], we can use a minimum of the following two.
a) Minimum of range [0, 3]
b) Minimum of range [4, 7]
Based on the above example, below is the formula,
// If arr[lookup[0][2]] <= arr[lookup[4][2]],
// then lookup[0][3] = lookup[0][2]
If arr[lookup[i][j-1]] <= arr[lookup[i+2j-1][j-1]]
lookup[i][j] = lookup[i][j-1]// If arr[lookup[0][2]] > arr[lookup[4][2]],
// then lookup[0][3] = lookup[4][2]
Else
lookup[i][j] = lookup[i+2j-1][j-1]
Query:
For any arbitrary range [l, R], we need to use ranges that are in powers of 2. The idea is to use the closest power of 2. We always need to do at most one comparison (compare a minimum of two ranges which are powers of 2). One range starts with L 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 the given range is (2, 10), we compare a minimum of two ranges (2, 9) and (3, 10).
Based on the above example, below is the formula,
// For (2,10), j = floor(Log2(10-2+1)) = 3
j = floor(Log(R-L+1))// If arr[lookup[0][3]] <= arr[lookup[3][3]],
// then RMQ(2,10) = lookup[0][3]
If arr[lookup[L][j]] <= arr[lookup[R-(int)pow(2,j)+1][j]]
RMQ(L, R) = lookup[L][j]// If arr[lookup[0][3]] > arr[lookup[3][3]],
// then RMQ(2,10) = lookup[3][3]
Else
RMQ(L, R) = lookup[R-(int)pow(2,j)+1][j]
Since we do only one comparison, the time complexity of the query is O(1).
Below is the implementation of the above idea.
C++
// C++ program to do range minimum
// query in O(1) time with
// O(n Log n) extra space and
// O(n Log n) preprocessing time
#include <bits/stdc++.h>
using namespace std;
#define MAX 500
// 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 range
struct 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][3]]
// and arr[lookup[3][3]]
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][3]] and
// arr[lookup[3][3]],
if (arr[lookup[L][j]]
<= arr[lookup[R - (1 << j) + 1][j]])
return arr[lookup[L][j]];
else
return arr[lookup[R - (1 << 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;
}
}
// Driver code
int main()
{
int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = sizeof(a) / sizeof(a[0]);
Query q[] = { { 0, 4 }, { 4, 7 }, { 7, 8 } };
int m = sizeof(q) / sizeof(q[0]);
RMQ(a, n, q, m);
return 0;
}
// Java program to do range minimum query
// in O(1) time with O(n Log n) extra space
// and O(n Log n) preprocessing time
import java.util.*;
class GFG {
static int MAX = 500;
// 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.
static int[][] lookup = new int[MAX][MAX];
// Structure to represent a query range
static class Query {
int L, R;
public Query(int L, int R)
{
this.L = L;
this.R = R;
}
};
// Fills lookup array lookup[][]
// in bottom up manner.
static 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][3]]
// and arr[lookup[3][3]]
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]
static int query(int arr[], int L, int R)
{
// For [2,10], j = 3
int j = (int)(Math.log(R - L + 1) / Math.log(2));
// For [2,10], we compare
// arr[lookup[0][3]]
// and arr[lookup[3][3]],
if (arr[lookup[L][j]]
<= arr[lookup[R - (1 << j) + 1][j]])
return arr[lookup[L][j]];
else
return arr[lookup[R - (1 << j) + 1][j]];
}
// Prints minimum of given m
// query ranges in arr[0..n-1]
static 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
System.out.println("Minimum of ["
+ L + ", " + R
+ "] is "
+ query(arr, L, R));
}
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = a.length;
Query q[] = { new Query(0, 4), new Query(4, 7),
new Query(7, 8) };
int m = q.length;
RMQ(a, n, q, m);
}
}
// This code is contributed by Rajput-Ji
# Python3 program to do range minimum query
# in O(1) time with O(n Log n) extra space
# and O(n Log n) preprocessing time
from math import log2
MAX = 500
# 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.
lookup = [[0 for i in range(500)]
for j in range(500)]
# Structure to represent a query range
class Query:
def __init__(self, l, r):
self.L = l
self.R = r
# Fills lookup array lookup[][]
# in bottom up manner.
def preprocess(arr: list, n: int):
global lookup
# Initialize M for the
# intervals with length 1
for i in range(n):
lookup[i][0] = i
# Compute values from
# smaller to bigger intervals
j = 1
while (1 << j) <= n:
# Compute minimum value for
# all intervals with size 2^j
i = 0
while i + (1 << j) - 1 < n:
# For arr[2][10], we compare
# arr[lookup[0][3]] and
# arr[lookup[3][3]]
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]
i += 1
j += 1
# Returns minimum of arr[L..R]
def query(arr: list, L: int, R: int) -> int:
global lookup
# For [2,10], j = 3
j = int(log2(R - L + 1))
# For [2,10], we compare
# arr[lookup[0][3]] and
# arr[lookup[3][3]],
if (arr[lookup[L][j]] <=
arr[lookup[R - (1 << j) + 1][j]]):
return arr[lookup[L][j]]
else:
return arr[lookup[R - (1 << j) + 1][j]]
# Prints minimum of given
# m query ranges in arr[0..n-1]
def RMQ(arr: list, n: int, q: list, m: int):
# Fills table lookup[n][Log n]
preprocess(arr, n)
# One by one compute sum of all queries
for i in range(m):
# Left and right boundaries
# of current range
L = q[i].L
R = q[i].R
# Print sum of current query range
print("Minimum of [%d, %d] is %d" %
(L, R, query(arr, L, R)))
# Driver Code
if __name__ == "__main__":
a = [7, 2, 3, 0, 5, 10, 3, 12, 18]
n = len(a)
q = [Query(0, 4), Query(4, 7),
Query(7, 8)]
m = len(q)
RMQ(a, n, q, m)
# This code is contributed by
# sanjeev2552
// C# program to do range minimum query
// in O(1) time with O(n Log n) extra space
// and O(n Log n) preprocessing time
using System;
class GFG {
static int MAX = 500;
// 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.
static int[, ] lookup = new int[MAX, MAX];
// Structure to represent a query range
public class Query {
public int L, R;
public Query(int L, int R)
{
this.L = L;
this.R = R;
}
};
// Fills lookup array lookup[,]
// in bottom up manner.
static 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[0,3]] and arr[lookup[3,3]]
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]
static int query(int[] arr, int L, int R)
{
// For [2,10], j = 3
int j = (int)Math.Log(R - L + 1);
// For [2,10], we compare arr[lookup[0,3]]
// and arr[lookup[3,3]],
if (arr[lookup[L, j]]
<= arr[lookup[R - (1 << j) + 1, j]])
return arr[lookup[L, j]];
else
return arr[lookup[R - (1 << j) + 1, j]];
}
// Prints minimum of given m
// query ranges in arr[0..n-1]
static 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
Console.WriteLine("Minimum of [" + L + ", " + R
+ "] is " + query(arr, L, R));
}
}
// Driver Code
public static void Main(String[] args)
{
int[] a = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int n = a.Length;
Query[] q = { new Query(0, 4), new Query(4, 7),
new Query(7, 8) };
int m = q.Length;
RMQ(a, n, q, m);
}
}
// This code is contributed by Princi Singh
<script>
// Javascript program to do range minimum query
// in O(1) time with O(n Log n) extra space
// and O(n Log n) preprocessing time
const MAX = 500;
// 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.
let lookup = new Array(MAX).fill(0).map(() => new Array(MAX))
// Structure to represent a query range
class Query {
constructor(L, R) {
this.L = L;
this.R = R;
}
};
// Fills lookup array lookup[][]
// in bottom up manner.
function preprocess(arr, n) {
// Initialize M for the intervals
// with length 1
for (let i = 0; i < n; i++)
lookup[i][0] = i;
// Compute values from smaller
// to bigger intervals
for (let j = 1; (1 << j) <= n; j++) {
// Compute minimum value for
// all intervals with size 2^j
for (let i = 0;
(i + (1 << j) - 1) < n;
i++) {
// For arr[2][10], we compare
// arr[lookup[0][3]]
// and arr[lookup[3][3]]
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]
function query(arr, L, R) {
// For [2,10], j = 3
let j = Math.floor(Math.log(R - L + 1));
// For [2,10], we compare
// arr[lookup[0][3]]
// and arr[lookup[3][3]],
if (arr[lookup[L][j]]
<= arr[lookup[R - (1 << j) + 1][j]])
return arr[lookup[L][j]];
else
return arr[lookup[R - (1 << j) + 1][j]];
}
// Prints minimum of given m
// query ranges in arr[0..n-1]
function RMQ(arr, n, q, m) {
// Fills table lookup[n][Log n]
preprocess(arr, n);
// One by one compute sum of all queries
for (let i = 0; i < m; i++) {
// Left and right boundaries
// of current range
let L = q[i].L, R = q[i].R;
// Print sum of current query range
document.write("Minimum of ["
+ L + ", " + R
+ "] is "
+ query(arr, L, R) + "<br>");
}
}
// Driver Code
let a = [7, 2, 3, 0, 5, 10, 3, 12, 18];
let n = a.length;
let q = [new Query(0, 4), new Query(4, 7),
new Query(7, 8)];
let m = q.length;
RMQ(a, n, q, m);
// This code is contributed by Saurabh jaiswal
</script>
OutputMinimum of [0, 4] is 0
Minimum of [4, 7] is 3
Minimum of [7, 8] is 12
So sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time and O(n Log n) space.
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem