Introduction and Dynamic Programming solution to compute nCr%p
Last Updated :
23 Jul, 2025
Given three numbers n, r and p, compute value of nCr mod p.
Example:
Input: n = 10, r = 2, p = 13
Output: 6
Explanation: 10C2 is 45 and 45 % 13 is 6.
METHOD 1: (Using Dynamic Programming)
A Simple Solution is to first compute nCr, then compute nCr % p. This solution works fine when the value of nCr is small.
What if the value of nCr is large?
The value of nCr%p is generally needed for large values of n when nCr cannot fit in a variable, and causes overflow. So computing nCr and then using modular operator is not a good idea as there will be overflow even for slightly larger values of n and r. For example the methods discussed here and here cause overflow for n = 50 and r = 40.
The idea is to compute nCr using below formula
C(n, r) = C(n-1, r-1) + C(n-1, r)
C(n, 0) = C(n, n) = 1
Working of Above formula and Pascal Triangle:
Let us see how above formula works for C(4, 3)
1==========>> n = 0, C(0, 0) = 1
1–1========>> n = 1, C(1, 0) = 1, C(1, 1) = 1
1–2–1======>> n = 2, C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1
1–3–3–1====>> n = 3, C(3, 0) = 1, C(3, 1) = 3, C(3, 2) = 3, C(3, 3)=1
1–4–6–4–1==>> n = 4, C(4, 0) = 1, C(4, 1) = 4, C(4, 2) = 6, C(4, 3)=4, C(4, 4)=1
So here every loop on i, builds i’th row of pascal triangle, using (i-1)th row
Extension of above formula for modular arithmetic:
We can use distributive property of modular operator to find nCr % p using above formula.
C(n, r)%p = [ C(n-1, r-1)%p + C(n-1, r)%p ] % p
C(n, 0) = C(n, n) = 1
The above formula can be implemented using Dynamic Programming using a 2D array.
The 2D array based dynamic programming solution can be further optimized by constructing one row at a time. See Space optimized version in below post for details.
Binomial Coefficient using Dynamic Programming
Below is implementation based on the space optimized version discussed in above post.
C++
// A Dynamic Programming based solution to compute nCr % p
#include <bits/stdc++.h>
using namespace std;
// Returns nCr % p
int nCrModp(int n, int r, int p)
{
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// The array C is going to store last row of
// pascal triangle at the end. And last entry
// of last row is nCr
int C[r + 1];
memset(C, 0, sizeof(C));
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows of Pascal
// Triangle from top to bottom
for (int i = 1; i <= n; i++) {
// Fill entries of current row using previous
// row values
for (int j = min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r];
}
// Driver program
int main()
{
int n = 10, r = 2, p = 13;
cout << "Value of nCr % p is " << nCrModp(n, r, p);
return 0;
}
// A Dynamic Programming based
// solution to compute nCr % p
import java.io.*;
import java.util.*;
import java.math.*;
class GFG {
// Returns nCr % p
static int nCrModp(int n, int r, int p)
{
if (r > n - r)
r = n - r;
// The array C is going to store last
// row of pascal triangle at the end.
// And last entry of last row is nCr
int C[] = new int[r + 1];
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows of Pascal
// Triangle from top to bottom
for (int i = 1; i <= n; i++) {
// Fill entries of current row using previous
// row values
for (int j = Math.min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r];
}
// Driver program
public static void main(String args[])
{
int n = 10, r = 2, p = 13;
System.out.println("Value of nCr % p is "
+ nCrModp(n, r, p));
}
}
// This code is contributed by Nikita Tiwari.
# A Dynamic Programming based solution to compute nCr % p
# Returns nCr % p
def nCrModp(n, r, p):
# Optimization for the cases when r is large
# compared to n-r
if (r > n- r):
r = n - r
# The array C is going to store last row of
# pascal triangle at the end. And last entry
# of last row is nCr.
C = [0 for i in range(r + 1)]
C[0] = 1 # Top row of Pascal Triangle
# One by constructs remaining rows of Pascal
# Triangle from top to bottom
for i in range(1, n + 1):
# Fill entries of current row
# using previous row values
for j in range(min(i, r), 0, -1):
# nCj = (n - 1)Cj + (n - 1)C(j - 1)
C[j] = (C[j] + C[j-1]) % p
return C[r]
# Driver Program
n = 10
r = 2
p = 13
print('Value of nCr % p is', nCrModp(n, r, p))
# This code is contributed by Soumen Ghosh
// A Dynamic Programming based
// solution to compute nCr % p
using System;
class GFG {
// Returns nCr % p
static int nCrModp(int n, int r, int p)
{
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// The array C is going to store last
// row of pascal triangle at the end.
// And last entry of last row is nCr
int[] C = new int[r + 1];
for (int i = 0; i < r + 1; i++)
C[i] = 0;
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows
// of Pascal Triangle from top to bottom
for (int i = 1; i <= n; i++) {
// Fill entries of current row using
// previous row values
for (int j = Math.Min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r];
}
// Driver program
public static void Main()
{
int n = 10, r = 2, p = 13;
Console.Write("Value of nCr % p is "
+ nCrModp(n, r, p));
}
}
// This code is contributed by nitin mittal.
<?php
// A Dynamic Programming based
// solution to compute nCr % p
// Returns nCr % p
function nCrModp($n, $r, $p)
{
// Optimization for the cases when r is large
if ($r > $n - $r)
$r = $n - $r;
// The array C is going
// to store last row of
// pascal triangle at
// the end. And last entry
// of last row is nCr
$C = array();
for( $i = 0; $i < $r + 1; $i++)
$C[$i] = 0;
// Top row of Pascal
// Triangle
$C[0] = 1;
// One by constructs remaining
// rows of Pascal Triangle from
// top to bottom
for ($i = 1; $i <= $n; $i++)
{
// Fill entries of current
// row using previous row values
for ($j = Min($i, $r); $j > 0; $j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
$C[$j] = ($C[$j] +
$C[$j - 1]) % $p;
}
return $C[$r];
}
// Driver Code
$n = 10; $r = 2;$p = 13;
echo "Value of nCr % p is ",
nCrModp($n, $r, $p);
// This code is contributed
// by anuj_67.
?>
<script>
// A Dynamic Programming based
// solution to compute nCr % p
// Returns nCr % p
function nCrModp(n,r,p)
{
if (r > n - r)
r = n - r;
// The array C is going to store last
// row of pascal triangle at the end.
// And last entry of last row is nCr
let C = new Array(r + 1);
for(let i = 0; i < r + 1; i++)
C[i] = 0;
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows of Pascal
// Triangle from top to bottom
for (let i = 1; i <= n; i++) {
// Fill entries of current row using previous
// row values
for (let j = Math.min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r];
}
// Driver program
let n = 10, r = 2, p = 13;
document.write("Value of nCr % p is "
+ nCrModp(n, r, p));
// This code is contributed by avanitrachhadiya2155
</script>
OutputValue of nCr % p is 6
Time complexity of above solution is O(n*r) and it requires O(r) space. There are more and better solutions to above problem.
Compute nCr % p | Set 2 (Lucas Theorem)
METHOD 2(Using Pascal Triangle and Dynamic Pro)
Another approach lies in utilizing the concept of the Pascal Triangle. Instead of calculating the nCr value for every n starting from n=0 till n=n, the approach aims at using the nth row itself for the calculation. The method proceeds by finding out a general relationship between nCr and nCr-1.
FORMULA: C(n,r)=C(n,r-1)* (n-r+1)/r
Example:
For instance, take n=5 and r=3.
Input: n = 5, r = 3, p = 1000000007
Output: 6
Explanation: 5C3 is 10 and 10 % 100000007 is 10.
As per the formula,
C(5,3)=5!/(3!)*(2!)
C(5,3)=10
Also,
C(5,2)=5!/(2!)*(3!)
C(5,2)=10
Let's try applying the above formula.
C(n,r)=C(n,r-1)* (n-r+1)/r
C(5,3)=C(5,2)*(5-3+1)/3
C(5,3)=C(5,2)*1
C(5,3)=10*1
The above example shows that C(n,r) can be easily calculated by calculating C(n,r-1) and multiplying the result with the term (n-r+1)/r. But this multiplication may cause integer overflow for large values of n. To tackle this situation, use modulo multiplication, and modulo division concepts in order to achieve optimizations in terms of integer overflow.
Let's find out how to build Pascal Triangle for the same.
{1}\\ {1\hspace{0.1cm} 1}\\ {1\hspace{0.1cm} 2\hspace{0.1cm} 1}\\ {1\hspace{0.1cm} 3\hspace{0.1cm} 3\hspace{0.1cm} 1}\\ {1 \hspace{0.1cm}4\hspace{0.1cm} 6\hspace{0.1cm} 4\hspace{0.1cm} 1}\\ {1\hspace{0.1cm} 5\hspace{0.1cm} 10\hspace{0.1cm} 10\hspace{0.1cm} 5\hspace{0.1cm} 1}
1D array declaration can be further optimized by just the declaration of a single variable to perform calculations. However, integer overflow demands other functions too for the final implementation.
The post below mentions the space and time-optimized implementation for the binary coefficient calculation.
C++
// C++ program to find the nCr%p
// based on optimal Dynamic
// Programming implementation and
// Pascal Triangle concepts
#include <bits/stdc++.h>
using namespace std;
// Returns (a * b) % mod
long long moduloMultiplication(long long a, long long b,
long long mod)
{
// Initialize result
long long res = 0;
// Update a if it is more than
// or equal to mod
a %= mod;
while (b) {
// If b is odd, add a with result
if (b & 1)
res = (res + a) % mod;
// Here we assume that doing 2*a
// doesn't cause overflow
a = (2 * a) % mod;
b >>= 1; // b = b / 2
}
return res;
}
// C++ function for extended Euclidean Algorithm
long long int gcdExtended(long long int a, long long int b,
long long int* x,
long long int* y);
// Function to find modulo inverse of b. It returns
// -1 when inverse doesn't exists
long long int modInverse(long long int b, long long int m)
{
long long int x, y; // used in extended GCD algorithm
long long int g = gcdExtended(b, m, &x, &y);
// Return -1 if b and m are not co-prime
if (g != 1)
return -1;
// m is added to handle negative x
return (x % m + m) % m;
}
// C++ function for extended Euclidean Algorithm (used to
// find modular inverse.
long long int gcdExtended(long long int a, long long int b,
long long int* x,
long long int* y)
{
// Base Case
if (a == 0) {
*x = 0, *y = 1;
return b;
}
// To store results of recursive call
long long int x1, y1;
long long int gcd = gcdExtended(b % a, a, &x1, &y1);
// Update x and y using results of recursive
// call
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
// Function to compute a/b under modulo m
long long int modDivide(long long int a, long long int b,
long long int m)
{
a = a % m;
long long int inv = modInverse(b, m);
if (inv == -1)
// cout << "Division not defined";
return 0;
else
return (inv * a) % m;
}
// Function to calculate nCr % p
int nCr(int n, int r, int p)
{
// Edge Case which is not possible
if (r > n)
return 0;
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// x stores the current result at
long long int x = 1;
// each iteration
// Initialized to 1 as nC0 is always 1.
for (int i = 1; i <= r; i++) {
// Formula derived for calculating result is
// C(n,r-1)*(n-r+1)/r
// Function calculates x*(n-i+1) % p.
x = moduloMultiplication(x, (n + 1 - i), p);
// Function calculates x/i % p.
x = modDivide(x, i, p);
}
return x;
}
// Driver Code
int main()
{
long long int n = 5, r = 3, p = 1000000007;
cout << "Value of nCr % p is " << nCr(n, r, p);
return 0;
}
// Java program to find the nCr%p
// based on optimal Dynamic
// Programming implementation and
// Pascal Triangle concepts
import java.util.*;
class GFG
{
// Returns (a * b) % mod
static long moduloMultiplication(long a, long b,
long mod)
{
// Initialize result
long res = 0;
// Update a if it is more than
// or equal to mod
a %= mod;
while (b > 0) {
// If b is odd, add a with result
if ((b & 1) != 0)
res = (res + a) % mod;
// Here we assume that doing 2*a
// doesn't cause overflow
a = (2 * a) % mod;
b >>= 1; // b = b / 2
}
return res;
}
// Global Variables
static long x, y;
// Function for extended Euclidean Algorithm
static long gcdExtended(long a, long b)
{
// Base Case
if (a == 0) {
x = 0;
y = 1;
return b;
}
// To store results of recursive call
long gcd = gcdExtended(b % a, a);
long x1 = x;
long y1 = y;
// Update x and y using results of recursive
// call
x = y1 - (b / a) * x1;
y = x1;
return gcd;
}
static long modInverse(long a, long m)
{
long g = gcdExtended(a, m);
// Return -1 if b and m are not co-prime
if (g != 1)
return -1;
// m is added to handle negative x
return (x % m + m) % m;
}
// Function to compute a/b under modulo m
static long modDivide(long a, long b, long m)
{
a = a % m;
long inv = modInverse(b, m);
if (inv == -1)
return 0;
else
return (inv * a) % m;
}
// Function to calculate nCr % p
static long nCr(long n, long r, long p)
{
// Edge Case which is not possible
if (r > n)
return 0;
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// x stores the current result at
long x = 1;
// each iteration
// Initialized to 1 as nC0 is always 1.
for (long i = 1L; i <= r; i++) {
// Formula derived for calculating result is
// C(n,r-1)*(n-r+1)/r
// Function calculates x*(n-i+1) % p.
x = moduloMultiplication(x, (n + 1L - i), p);
// Function calculates x/i % p.
x = modDivide(x, i, p);
}
return x;
}
// Driver Code
public static void main(String[] args)
{
long n = 5, r = 3, p = 1000000007;
System.out.println("Value of nCr % p is "
+ nCr(n, r, p));
}
}
// This code is contributed by phasing17
# Python3 program to find the nCr%p
# based on optimal Dynamic
# Programming implementation and
# Pascal Triangle concepts
# Returns (a * b) % mod
def moduloMultiplication(a, b, mod):
# Initialize result
res = 0
# Update a if it is more than
# or equal to mod
a %= mod
while (b):
# If b is odd, add a with result
if (b & 1):
res = (res + a) % mod
# Here we assume that doing 2*a
# doesn't cause overflow
a = (2 * a) % mod
b >>= 1 # b = b / 2
return res
# Global Variables
x, y = 0, 1
# Function for extended Euclidean Algorithm
def gcdExtended(a, b):
global x, y
# Base Case
if (a == 0):
x = 0
y = 1
return b
# To store results of recursive call
gcd = gcdExtended(b % a, a)
x1 = x
y1 = y
# Update x and y using results of recursive
# call
x = y1 - int(b / a) * x1
y = x1
return gcd
def modInverse(a, m):
g = gcdExtended(a, m)
# Return -1 if b and m are not co-prime
if (g != 1):
return -1
# m is added to handle negative x
return (x % m + m) % m
# Function to compute a/b under modulo m
def modDivide(a, b, m):
a = a % m
inv = modInverse(b, m)
if (inv == -1):
return 0
else:
return (inv * a) % m
# Function to calculate nCr % p
def nCr(n, r, p):
# Edge Case which is not possible
if (r > n):
return 0
# Optimization for the cases when r is large
if (r > n - r):
r = n - r
# x stores the current result at
x = 1
# each iteration
# Initialized to 1 as nC0 is always 1.
for i in range(1, r + 1):
# Formula derived for calculating result is
# C(n,r-1)*(n-r+1)/r
# Function calculates x*(n-i+1) % p.
x = moduloMultiplication(x, (n + 1 - i), p)
# Function calculates x/i % p.
x = modDivide(x, i, p)
return x
# Driver Code
n = 5
r = 3
p = 1000000007
print("Value of nCr % p is ", nCr(n, r, p))
# This code is contributed by phasing17
// C# program to find the nCr%p
// based on optimal Dynamic
// Programming implementation and
// Pascal Triangle concepts
using System;
using System.Collections.Generic;
class GFG
{
// Returns (a * b) % mod
static long moduloMultiplication(long a, long b, long mod)
{
// Initialize result
long res = 0;
// Update a if it is more than
// or equal to mod
a %= mod;
while (b > 0) {
// If b is odd, add a with result
if ((b & 1) != 0)
res = (res + a) % mod;
// Here we assume that doing 2*a
// doesn't cause overflow
a = (2 * a) % mod;
b >>= 1; // b = b / 2
}
return res;
}
// Global Variables
static long x, y;
// Function for extended Euclidean Algorithm
static long gcdExtended(long a, long b){
// Base Case
if (a == 0)
{
x = 0;
y = 1;
return b;
}
// To store results of recursive call
long gcd = gcdExtended(b % a, a);
long x1 = x;
long y1 = y;
// Update x and y using results of recursive
// call
x = y1 - (b / a) * x1;
y = x1;
return gcd;
}
static long modInverse(long a, long m)
{
long g = gcdExtended(a, m);
// Return -1 if b and m are not co-prime
if (g != 1)
return -1;
// m is added to handle negative x
return (x % m + m) % m;
}
// Function to compute a/b under modulo m
static long modDivide(long a, long b, long m)
{
a = a % m;
long inv = modInverse(b, m);
if (inv == -1)
return 0;
else
return (inv * a) % m;
}
// Function to calculate nCr % p
static long nCr(long n, long r, long p)
{
// Edge Case which is not possible
if (r > n)
return 0;
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// x stores the current result at
long x = 1;
// each iteration
// Initialized to 1 as nC0 is always 1.
for (long i = 1L; i <= r; i++) {
// Formula derived for calculating result is
// C(n,r-1)*(n-r+1)/r
// Function calculates x*(n-i+1) % p.
x = moduloMultiplication(x, (n + 1L - i), p);
// Function calculates x/i % p.
x = modDivide(x, i, p);
}
return x;
}
// Driver Code
public static void Main(string[] args)
{
long n = 5, r = 3, p = 1000000007;
Console.Write("Value of nCr % p is " + nCr(n, r, p));
}
}
// This code is contributed by phasing17
// JavaScript program to find the nCr%p
// based on optimal Dynamic
// Programming implementation and
// Pascal Triangle concepts
// Returns (a * b) % mod
function moduloMultiplication(a, b, mod)
{
// Initialize result
let res = 0;
// Update a if it is more than
// or equal to mod
a %= mod;
while (b) {
// If b is odd, add a with result
if (b & 1)
res = (res + a) % mod;
// Here we assume that doing 2*a
// doesn't cause overflow
a = (2 * a) % mod;
b >>= 1; // b = b / 2
}
return res;
}
// Global Variables
let x, y;
// Function for extended Euclidean Algorithm
function gcdExtended(a, b){
// Base Case
if (a == 0)
{
x = 0;
y = 1;
return b;
}
// To store results of recursive call
let gcd = gcdExtended(b % a, a);
let x1 = x;
let y1 = y;
// Update x and y using results of recursive
// call
x = y1 - Math.floor(b / a) * x1;
y = x1;
return gcd;
}
function modInverse(a, m)
{
let g = gcdExtended(a, m);
// Return -1 if b and m are not co-prime
if (g != 1)
return -1;
// m is added to handle negative x
return (x % m + m) % m;
}
// Function to compute a/b under modulo m
function modDivide(a, b, m)
{
a = a % m;
let inv = modInverse(b, m);
if (inv == -1)
return 0;
else
return (inv * a) % m;
}
// Function to calculate nCr % p
function nCr(n, r, p)
{
// Edge Case which is not possible
if (r > n)
return 0;
// Optimization for the cases when r is large
if (r > n - r)
r = n - r;
// x stores the current result at
let x = 1;
// each iteration
// Initialized to 1 as nC0 is always 1.
for (var i = 1; i <= r; i++) {
// Formula derived for calculating result is
// C(n,r-1)*(n-r+1)/r
// Function calculates x*(n-i+1) % p.
x = moduloMultiplication(x, (n + 1 - i), p);
// Function calculates x/i % p.
x = modDivide(x, i, p);
}
return x;
}
// Driver Code
let n = 5, r = 3, p = 1000000007;
console.log("Value of nCr % p is ", nCr(n, r, p));
// This code is contributed by phasing17
OutputValue of nCr % p is 10
Complexity Analysis:
- The above code needs an extra of O(1) space for the calculations.
- The time involved in the calculation of nCr % p is of the order O(n).
This article is improved by Aryan Gupta.
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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
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
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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
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
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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
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
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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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