Sparse table concept is used for fast queries on a set of static data (elements do not change). It does preprocessing so that the queries can be answered efficiently.
Range Minimum Query Using Sparse Table
You are given an integer array arr of length n and an integer q denoting the number of queries. Each query consists of two indices L and R (0 ≤ L ≤ R < n), and asks for the minimum value in the subarray arr[L…R].
Example:
Input: arr[] = [ 7, 2, 3, 0, 5, 10, 3, 12, 18 ]
queries[][] = [ [0, 4], [4, 7], [7, 8] ]
Output: 0 3 12
Explanation: For query 1, the subarray spanning indices 0 through 4 contains the values 7, 2, 3, and 0, and the minimum among them is 0. Similarly, the minimum value in range [4, 7] and [7, 8] are 3 and 12 respectively.
Approach:
The idea is to precompute the minimum values for all subarrays whose lengths are powers of two and store them in a table so that any range-minimum query can be answered in constant time. We build a 2D array lookup where lookup[i][j] holds the minimum of the subarray starting at i with length 2^j (j varies from to Log n where n is the length of the input array). For example lookup[0][3] contains minimum of range [0, 7] (starting with 0 and of size 23)
How to fill this lookup or sparse table?
The idea is simple, fill in a bottom-up manner using previously computed values. We compute ranges with current power of 2 using values of lower power of two. Each entry for length 2^j is derived by combining two overlapping subarrays of length 2^(j–1) that were computed in the previous step. For example, to find a minimum of range [0, 7] (Range size is a power of 3), we can use the minimum of following two.
a) Minimum of range [0, 3] (Range size is a power of 2)
b) Minimum of range [4, 7] (Range size is a power of 2)
Based on above example, below is formula,
// Minimum of single element subarrays is same
// as the only element.
lookup[i][0] = arr[i]
// If lookup[0][2] <= lookup[4][2],
// then lookup[0][3] = lookup[0][2]
If lookup[i][j-1] <= lookup[i+2j-1][j-1]
lookup[i][j] = lookup[i][j-1]
// If lookup[0][2] > lookup[4][2],
// then lookup[0][3] = lookup[4][2]
Else
lookup[i][j] = lookup[i+2j-1][j-1]
Follow the below given steps:
- Initialize
lookup[i][0] = arr[i] for every 0 ≤ i < n. - For each
j from 1 up to ⌊log₂n⌋, and for each i from 0 to n – 2^j:- Compute
lookup[i][j] as the minimum of the two halves:- the interval beginning at
i of length 2^(j–1) (lookup[i][j–1]) - the interval beginning at
i + 2^(j–1) of length 2^(j–1) (lookup[i + 2^(j–1)][j–1])
- To answer a query on the range
[L, R]:- Let
k = ⌊log₂(R – L + 1)⌋. - The minimum over
[L, R] is min(lookup[L][k], lookup[R – 2^k + 1][k]).
How do we handle individual queries?
- Find highest power of 2 that is smaller than or equal to count of elements in given range [L, R]. we get
j = floor(log2(R - L + 1)) For [2, 10], j = 3 - Compute minimum of last 2^j elements with first 2^j elements in range. For [2, 10], we compare lookup[0][3] (minimum from 0 to 7) and lookup[3][3] (minimum from 3 to 10) and return the minimum of two values.
C++
#include <bits/stdc++.h>
using namespace std;
// Fills lookup array lookup[][] in bottom up manner.
vector<vector<int>> buildSparseTable(vector<int> &arr) {
int n = arr.size();
// create the 2d table
vector<vector<int>> lookup(n + 1,
vector<int>(log2(n) + 1));
// Initialize for the intervals with length 1
for (int i = 0; i < n; i++)
lookup[i][0] = arr[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++) {
if (lookup[i][j - 1] <
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];
}
}
return lookup;
}
// Returns minimum of arr[L..R]
int query(int L, int R, vector<vector<int>> &lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements in range
int j = (int)log2(R - L + 1);
// Compute minimum of last 2^j elements with first
// 2^j elements in range.
if (lookup[L][j] <= lookup[R - (1 << j) + 1][j])
return lookup[L][j];
else
return lookup[R - (1 << j) + 1][j];
}
vector<int> solveQueries(vector<int>& arr,
vector<vector<int>>& queries) {
int n = arr.size();
int m = queries.size();
vector<int> result(m);
// Build the sparse table
vector<vector<int>> lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
int main() {
vector<int> arr = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
vector<vector<int>> queries = { {0, 4}, {4, 7}, {7, 8} };
vector<int> res = solveQueries(arr, queries);
for (int i = 0; i < res.size(); i++) {
cout << res[i] << " ";
}
return 0;
}
public class GfG {
// Fills lookup array lookup[][] in bottom up manner.
public static int[][] buildSparseTable(int[] arr) {
int n = arr.length;
// create the 2d table
int[][] lookup = new int[n + 1][(int)(Math.log(n) / Math.log(2)) + 1];
// Initialize for the intervals with length 1
for (int i = 0; i < n; i++)
lookup[i][0] = arr[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++) {
if (lookup[i][j - 1] <
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];
}
}
return lookup;
}
// Returns minimum of arr[L..R]
public static int query(int L, int R, int[][] lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements in range
int j = (int)(Math.log(R - L + 1) / Math.log(2));
// Compute minimum of last 2^j elements with first
// 2^j elements in range.
if (lookup[L][j] <= lookup[R - (1 << j) + 1][j])
return lookup[L][j];
else
return lookup[R - (1 << j) + 1][j];
}
public static int[] solveQueries(int[] arr, int[][] queries) {
int n = arr.length;
int m = queries.length;
int[] result = new int[m];
// Build the sparse table
int[][] lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
public static void main(String[] args) {
int[] arr = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int[][] queries = { {0, 4}, {4, 7}, {7, 8} };
int[] res = solveQueries(arr, queries);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
}
import math
# Fills lookup array lookup[][] in bottom up manner.
def buildSparseTable(arr):
n = len(arr)
# create the 2d table
lookup = [[0] * (int(math.log(n, 2)) + 1) for _ in range(n + 1)]
# Initialize for the intervals with length 1
for i in range(n):
lookup[i][0] = arr[i]
# Compute values from smaller to bigger intervals
for j in range(1, int(math.log(n, 2)) + 1):
# Compute minimum value for all intervals with
# size 2^j
for i in range(0, n - (1 << j) + 1):
if lookup[i][j - 1] < 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]
return lookup
# Returns minimum of arr[L..R]
def query(L, R, lookup):
# Find highest power of 2 that is smaller
# than or equal to count of elements in range
j = int(math.log(R - L + 1, 2))
# Compute minimum of last 2^j elements with first
# 2^j elements in range.
if lookup[L][j] <= lookup[R - (1 << j) + 1][j]:
return lookup[L][j]
else:
return lookup[R - (1 << j) + 1][j]
def solveQueries(arr, queries):
n = len(arr)
m = len(queries)
result = [0] * m
# Build the sparse table
lookup = buildSparseTable(arr)
# Process each query
for i in range(m):
L = queries[i][0]
R = queries[i][1]
result[i] = query(L, R, lookup)
return result
if __name__ == "__main__":
arr = [ 7, 2, 3, 0, 5, 10, 3, 12, 18 ]
queries = [ [0, 4], [4, 7], [7, 8] ]
res = solveQueries(arr, queries)
for x in res:
print(x, end=" ")
using System;
public class GfG {
// Fills lookup array lookup[][] in bottom up manner.
public static int[][] buildSparseTable(int[] arr) {
int n = arr.Length;
// create the 2d table
int cols = (int)(Math.Log(n) / Math.Log(2)) + 1;
int[][] lookup = new int[n + 1][];
for (int i = 0; i < n + 1; i++)
lookup[i] = new int[cols];
// Initialize for the intervals with length 1
for (int i = 0; i < n; i++)
lookup[i][0] = arr[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++) {
if (lookup[i][j - 1] <
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];
}
}
return lookup;
}
// Returns minimum of arr[L..R]
public static int query(int L, int R, int[][] lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements in range
int j = (int)(Math.Log(R - L + 1) / Math.Log(2));
// Compute minimum of last 2^j elements with first
// 2^j elements in range.
if (lookup[L][j] <= lookup[R - (1 << j) + 1][j])
return lookup[L][j];
else
return lookup[R - (1 << j) + 1][j];
}
public static int[] solveQueries(int[] arr, int[][] queries) {
int n = arr.Length;
int m = queries.Length;
int[] result = new int[m];
// Build the sparse table
int[][] lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
public static void Main(string[] args) {
int[] arr = { 7, 2, 3, 0, 5, 10, 3, 12, 18 };
int[][] queries = { new int[]{0, 4}, new int[]{4, 7}, new int[]{7, 8} };
int[] res = solveQueries(arr, queries);
for (int i = 0; i < res.Length; i++) {
Console.Write(res[i] + " ");
}
}
}
// Fills lookup array lookup[][] in bottom up manner.
function buildSparseTable(arr) {
const n = arr.length;
// create the 2d table
const cols = Math.floor(Math.log2(n)) + 1;
const lookup = Array.from({ length: n + 1 }, () => Array(cols).fill(0));
// Initialize for the intervals with length 1
for (let i = 0; i < n; i++)
lookup[i][0] = arr[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++) {
if (lookup[i][j - 1] <
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];
}
}
return lookup;
}
// Returns minimum of arr[L..R]
function query(L, R, lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements in range
const j = Math.floor(Math.log2(R - L + 1));
// Compute minimum of last 2^j elements with first
// 2^j elements in range.
if (lookup[L][j] <= lookup[R - (1 << j) + 1][j])
return lookup[L][j];
else
return lookup[R - (1 << j) + 1][j];
}
function solveQueries(arr, queries) {
const n = arr.length;
const m = queries.length;
const result = new Array(m);
// Build the sparse table
const lookup = buildSparseTable(arr);
// Process each query
for (let i = 0; i < m; i++) {
const L = queries[i][0];
const R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
const arr = [ 7, 2, 3, 0, 5, 10, 3, 12, 18 ];
const queries = [ [0, 4], [4, 7], [7, 8] ];
const res = solveQueries(arr, queries);
console.log(res.join(' '));
Time Complexity: O(n * log n), sparse table method supports query operation in O(1) time with O(n Log n) preprocessing time.
Auxiliary Space: O(n * log n)
Range GCD Query Using Sparse Table
You are given an integer array arr of length n and an integer q denoting the number of queries. Each query consists of two indices L and R (0 ≤ L ≤ R < n), and asks for the greatest common divisor in the subarray arr[L…R].
Example:
Input: arr[] = [2, 3, 5, 4, 6, 8]
queries[][] = [ [0, 2], [3, 5], [2, 3] ]
Output: 1 2 1
Explanation: For query 1, gcd of elements 2, 3, and 5 is 1 (as all three are primes).
For query 2, gcd of elements 4, 6, and 8 is 2.
For query 3, gcd of elements 5 and 4 is 1 (as 4 and 5 are co-primes).
Approach:
The idea is to exploit the associativity and idempotence of GCD to answer any range‐GCD query in constant time after preprocessing. Since
GCD(a, b, c) = GCD(GCD(a, b), c) = GCD(a, GCD(b, c))
and taking GCD of an overlapping element more than once does not change the result (e.g. GCD(a, b, c) = GCD(GCD(a, b), GCD(b, c))), we can build a sparse table over powers of two just like in the range‐minimum query. Any query range [L, R] can then be covered by two overlapping power‐of‐two intervals whose GCDs we combine to get the answer.
How to handle individual queries?
- Find highest power of 2 that is smaller than or equal to count of elements in given range [L, R]. we get
j = floor(log2(R - L + 1)) For [2, 10], j = 3 - Compute GCD of last 2^j elements with first 2^j elements in range. For [2, 10], we compute GCD of lookup[0][3] (GCD of 0 to 7) and GCD of lookup[3][3] (GCD of 3 to 10).
Follow the below given steps:
- For every index i, set lookup[i][0] = arr[i].
- For j = 1 to ⌊log₂n⌋, and for each i from 0 to n − 2ʲ:
- lookup[i][j] = GCD( lookup[i][j−1], lookup[i + 2^(j−1)][j−1] )
- To answer a query [L, R]:
- Let k = ⌊log₂(R − L + 1)⌋
- The GCD over [L, R] is GCD( lookup[L][k], lookup[R − 2ᵏ + 1][k] )
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Fills lookup array lookup[][] in bottom up manner.
vector<vector<int>> buildSparseTable(vector<int> &arr){
int n = arr.size();
// create the 2d table
vector<vector<int>> lookup(n + 1,
vector<int>(log2(n) + 1));
// GCD of single element is element itself
for (int i = 0; i < n; i++)
lookup[i][0] = arr[i];
// Build sparse table
for (int j = 1; j <= log2(n); j++)
for (int i = 0; i <= n - (1 << j); i++)
lookup[i][j] = __gcd(lookup[i][j - 1],
lookup[i + (1 << (j - 1))][j - 1]);
return lookup;
}
// Returns GCD of arr[L..R]
int query(int L, int R, vector<vector<int>> & lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements
int j = (int)log2(R - L + 1);
// Compute GCD of last 2^j elements with first
// 2^j elements in range.
return __gcd(lookup[L][j], lookup[R - (1 << j) + 1][j]);
}
vector<int> solveQueries(vector<int>& arr,
vector<vector<int>>& queries) {
int n = arr.size();
int m = queries.size();
vector<int> result(m);
// Build the sparse table
vector<vector<int>> lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
int main() {
vector<int> arr = { 2, 3, 5, 4, 6, 8 };
vector<vector<int>> queries = { {0, 2}, {3, 5}, {2, 3} };
vector<int> res = solveQueries(arr, queries);
for (int i = 0; i < res.size(); i++) {
cout << res[i] << " ";
}
return 0;
}
public class GfG {
public static int[][] buildSparseTable(int[] arr) {
int n = arr.length;
// create the 2d table
int[][] lookup = new int[n + 1][(int)(Math.log(n) / Math.log(2)) + 1];
// GCD of single element is element itself
for (int i = 0; i < n; i++)
lookup[i][0] = arr[i];
// Build sparse table
for (int j = 1; j <= (int)(Math.log(n) / Math.log(2)); j++)
for (int i = 0; i <= n - (1 << j); i++)
lookup[i][j] = gcd(lookup[i][j - 1],
lookup[i + (1 << (j - 1))][j - 1]);
return lookup;
}
public static int query(int L, int R, int[][] lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements
int j = (int)(Math.log(R - L + 1) / Math.log(2));
// Compute GCD of last 2^j elements with first
// 2^j elements in range.
return gcd(lookup[L][j], lookup[R - (1 << j) + 1][j]);
}
public static int[] solveQueries(int[] arr, int[][] queries) {
int n = arr.length;
int m = queries.length;
int[] result = new int[m];
// Build the sparse table
int[][] lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
private static int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
public static void main(String[] args) {
int[] arr = { 2, 3, 5, 4, 6, 8 };
int[][] queries = { {0, 2}, {3, 5}, {2, 3} };
int[] res = solveQueries(arr, queries);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
}
import math
def buildSparseTable(arr):
n = len(arr)
# create the 2d table
lookup = [[0] * (int(math.log2(n)) + 1) for _ in range(n + 1)]
# GCD of single element is element itself
for i in range(n):
lookup[i][0] = arr[i]
# Build sparse table
for j in range(1, int(math.log2(n)) + 1):
for i in range(0, n - (1 << j) + 1):
lookup[i][j] = math.gcd(lookup[i][j - 1],
lookup[i + (1 << (j - 1))][j - 1])
return lookup
def query(L, R, lookup):
# Find highest power of 2 that is smaller
# than or equal to count of elements
j = int(math.log2(R - L + 1))
# Compute GCD of last 2^j elements with first
# 2^j elements in range.
return math.gcd(lookup[L][j], lookup[R - (1 << j) + 1][j])
def solveQueries(arr, queries):
n = len(arr)
m = len(queries)
result = [0] * m
# Build the sparse table
lookup = buildSparseTable(arr)
# Process each query
for i in range(m):
L = queries[i][0]
R = queries[i][1]
result[i] = query(L, R, lookup)
return result
if __name__ == "__main__":
arr = [ 2, 3, 5, 4, 6, 8 ]
queries = [ [0, 2], [3, 5], [2, 3] ]
res = solveQueries(arr, queries)
for i in range(len(res)):
print(res[i], end=" ")
using System;
public class GfG {
public static int[][] buildSparseTable(int[] arr) {
int n = arr.Length;
// create the 2d table
int[][] lookup = new int[n + 1][];
for (int i = 0; i <= n; i++)
lookup[i] = new int[(int)(Math.Log(n) / Math.Log(2)) + 1];
// GCD of single element is element itself
for (int i = 0; i < n; i++)
lookup[i][0] = arr[i];
// Build sparse table
for (int j = 1; j <= (int)(Math.Log(n) / Math.Log(2)); j++)
for (int i = 0; i <= n - (1 << j); i++)
lookup[i][j] = GCD(lookup[i][j - 1],
lookup[i + (1 << (j - 1))][j - 1]);
return lookup;
}
public static int query(int L, int R, int[][] lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements
int j = (int)(Math.Log(R - L + 1) / Math.Log(2));
// Compute GCD of last 2^j elements with first
// 2^j elements in range.
return GCD(lookup[L][j], lookup[R - (1 << j) + 1][j]);
}
public static int[] solveQueries(int[] arr, int[][] queries) {
int n = arr.Length;
int m = queries.Length;
int[] result = new int[m];
// Build the sparse table
int[][] lookup = buildSparseTable(arr);
// Process each query
for (int i = 0; i < m; i++) {
int L = queries[i][0];
int R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
private static int GCD(int a, int b) {
return b == 0 ? a : GCD(b, a % b);
}
public static void Main(string[] args) {
int[] arr = { 2, 3, 5, 4, 6, 8 };
int[][] queries = new int[][] { new int[]{0, 2},
new int[]{3, 5}, new int[]{2, 3} };
int[] res = solveQueries(arr, queries);
for (int i = 0; i < res.Length; i++) {
Console.Write(res[i] + " ");
}
}
}
// Fills lookup array lookup[][] in bottom up manner.
function buildSparseTable(arr) {
const n = arr.length;
// create the 2d table
const lookup = Array.from({ length: n + 1 },
() => Array(Math.floor(Math.log2(n)) + 1).fill(0));
// GCD of single element is element itself
for (let i = 0; i < n; i++)
lookup[i][0] = arr[i];
// Build sparse table
for (let j = 1; j <= Math.floor(Math.log2(n)); j++)
for (let i = 0; i <= n - (1 << j); i++)
lookup[i][j] = gcd(lookup[i][j - 1],
lookup[i + (1 << (j - 1))][j - 1]);
return lookup;
}
// Returns GCD of arr[L..R]
function query(L, R, lookup) {
// Find highest power of 2 that is smaller
// than or equal to count of elements
const j = Math.floor(Math.log2(R - L + 1));
// Compute GCD of last 2^j elements with first
// 2^j elements in range.
return gcd(lookup[L][j], lookup[R - (1 << j) + 1][j]);
}
function solveQueries(arr, queries) {
const n = arr.length;
const m = queries.length;
const result = new Array(m);
// Build the sparse table
const lookup = buildSparseTable(arr);
// Process each query
for (let i = 0; i < m; i++) {
const L = queries[i][0];
const R = queries[i][1];
result[i] = query(L, R, lookup);
}
return result;
}
function gcd(a, b) {
return b === 0 ? a : gcd(b, a % b);
}
const arr = [ 2, 3, 5, 4, 6, 8 ];
const queries = [ [0, 2], [3, 5], [2, 3] ];
const res = solveQueries(arr, queries);
console.log(res.join(" "));
Time Complexity: O(n * log n)
Auxiliary Space: O(n * log n)
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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
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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
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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
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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:
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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