Sum of length of two smallest subsets possible from a given array with sum at least K
Last Updated :
26 Jul, 2025
Given an array arr[] consisting of N integers and an integer K, the task is to find the sum of the length of the two smallest unique subsets having sum of its elements at least K.
Examples:
Input: arr[] = {2, 4, 5, 6, 7, 8}, K = 16
Output: 6
Explanation:
The subsets {2, 6, 8} and {4, 5, 7} are the two smallest subsets with sum K(= 16).
Therefore, the sum of the lengths of both these subsets = 3 + 3 = 6.
Input: arr[] = {14, 3, 7, 8, 9, 7, 12, 15, 10, 6}, K = 40
Output: 8
Approach: The given problem can be solved based on the following observations:
- Sorting the array reduces the problem to choosing a subarray whose sum is at least K between the range of indices [i, N], and then check, if the sum of the remaining array elements in the range of indices [i, N] is K or not.
- To implement the above idea, a 2D array, say dp[][], is used such that dp[i][j] stores the minimum sum of the subset over the range of indices [i, N] having a value at least j. Then the transition state is similar to 0/1 Knapsack that can be defined as:
- If the value of arr[i] is greater than j, then update dp[i][j] to arr[i].
- Otherwise, update dp[i][j] to the minimum of dp[i + 1][j] and (dp[i + 1][j - arr[i]] + arr[i]).
Follow the steps below to solve the problem:
- Sort the array in ascending order.
- Initialize an array, say suffix[], and store the suffix sum of the array arr[] in it.
- Initialize a 2D array, say dp[][], such that dp[i][j] stores the minimum sum of the subset over the range of indices [i, N] having a value at least j.
- Initialize dp[N][0] as 0 and all other states as INT_MAX.
- Traverse the array arr[i] in reverse order and perform the following steps:
- Iterate over the range of indices [0, K] in reverse order and perform the following operations:
- If the value of arr[i] is at least j, then update the value of dp[i][j] as arr[i] as the current state has sum at least j. Now, continue the iteration.
- If the value of next state, i.e., dp[i + 1][j - arr[i]] is INT_MAX, then update dp[i][j] as INT_MAX.
- Otherwise, update dp[i][j] as the minimum of dp[i + 1][j] and (dp[i + 1][j - arr[i]] + arr[i]) to store the sum of all values having sum at least j.
- Now, traverse the array suffix[] in reverse order and if the value of (suffix[i] - dp[i][K]) is at least K, then print (N - i) as the sum of the size of the two smallest subsets formed and break out of the loop.
- Otherwise, print "-1".
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e9;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
int MinimumLength(int A[], int N, int K)
{
// Sort the array in ascending order
sort(A, A + N);
// Stores suffix sum of the array
int suffix[N + 1] = { 0 };
// Update the suffix sum array
for (int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Stores all dp-states
int dp[N + 1][K + 1];
// Initialize all dp-states
// with a maximum possible value
for (int i = 0; i <= N; i++)
for (int j = 0; j <= K; j++)
dp[i][j] = MAX;
// Base Case
dp[N][0] = 0;
// Traverse the array arr[]
for (int i = N - 1; i >= 0; i--) {
// Iterate over the range [0, K]
for (int j = K; j >= 0; j--) {
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i]) {
dp[i][j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (dp[i + 1][j - A[i]] == MAX)
dp[i][j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
dp[i][j] = min(dp[i + 1][j],
dp[i + 1][j - A[i]] + A[i]);
}
}
// Traverse the suffix sum array
for (int i = N - 1; i >= 0; i--) {
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - dp[i][K] >= K) {
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
int main()
{
int arr[] = { 7, 4, 5, 6, 8 };
int K = 13;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MinimumLength(arr, N, K);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
static int MAX = (int)(1e9);
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int A[], int N, int K)
{
// Sort the array in ascending order
Arrays.sort(A);
// Stores suffix sum of the array
int suffix[] = new int[N + 1];
// Update the suffix sum array
for(int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Stores all dp-states
int dp[][] = new int[N + 1][K + 1];
// Initialize all dp-states
// with a maximum possible value
for(int i = 0; i <= N; i++)
for(int j = 0; j <= K; j++)
dp[i][j] = MAX;
// Base Case
dp[N][0] = 0;
// Traverse the array arr[]
for(int i = N - 1; i >= 0; i--)
{
// Iterate over the range [0, K]
for(int j = K; j >= 0; j--)
{
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i])
{
dp[i][j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (dp[i + 1][j - A[i]] == MAX)
dp[i][j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
dp[i][j] = Math.min(dp[i + 1][j],
dp[i + 1][j - A[i]]
+ A[i]);
}
}
// Traverse the suffix sum array
for(int i = N - 1; i >= 0; i--)
{
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - dp[i][K] >= K)
{
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 7, 4, 5, 6, 8 };
int K = 13;
int N = arr.length;
System.out.println(MinimumLength(arr, N, K));
}
}
// This code is contributed by Kingash
# Python3 program for the above approach
MAX = 1e9
# Function to calculate sum of lengths
# of two smallest subsets with sum >= K
def MinimumLength(A, N, K):
# Sort the array in ascending order
A.sort()
# Stores suffix sum of the array
suffix = [0] * (N + 1)
# Update the suffix sum array
for i in range(N - 1, -1, -1):
suffix[i] = suffix[i + 1] + A[i]
# Stores all dp-states
dp = [[0] * (K + 1)] * (N + 1)
# Initialize all dp-states
# with a maximum possible value
for i in range(N + 1):
for j in range(K + 1):
dp[i][j] = MAX
# Base Case
dp[N][0] = 0
# Traverse the array arr[]
for i in range(N - 1, -1, -1):
# Iterate over the range [0, K]
for j in range(K, -1, -1):
# If A[i] is equal to at
# least the required sum
# j for the current state
if (j <= A[i]) :
dp[i][j] = A[i]
continue
# If the next possible
# state doesn't exist
if (dp[i + 1][j - A[i]] == MAX):
dp[i][j] = MAX
# Otherwise, update the current
# state to the minimum of the
# next state and state including
# the current element A[i]
else :
dp[i][j] = min(dp[i + 1][j],
dp[i + 1][j - A[i]] + A[i])
# Traverse the suffix sum array
for i in range(N - 1, -1, -1):
# If suffix[i] - dp[i][K] >= K
if (suffix[i] - dp[i][K] >= K):
# Sum of lengths of the two
# smallest subsets is obtained
return N - i
# Return -1, if there doesn't
# exist any subset of sum >= K
return -1
# Driver Code
arr = [ 7, 4, 5, 6, 8 ]
K = 13
N = len(arr)
print(MinimumLength(arr, N, K))
# This code is contributed by splevel62
// C# program for the above approach
using System;
class GFG{
static int MAX = (int)(1e9);
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int[] A, int N, int K)
{
// Sort the array in ascending order
Array.Sort(A);
// Stores suffix sum of the array
int[] suffix = new int[N + 1];
// Update the suffix sum array
for(int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Stores all dp-states
int[,] dp = new int[N + 1, K + 1];
// Initialize all dp-states
// with a maximum possible value
for(int i = 0; i <= N; i++)
for(int j = 0; j <= K; j++)
dp[i, j] = MAX;
// Base Case
dp[N, 0] = 0;
// Traverse the array arr[]
for(int i = N - 1; i >= 0; i--)
{
// Iterate over the range [0, K]
for(int j = K; j >= 0; j--)
{
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i])
{
dp[i, j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (dp[i + 1, j - A[i]] == MAX)
dp[i, j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
dp[i, j] = Math.Min(dp[i + 1, j],
dp[i + 1, j - A[i]]
+ A[i]);
}
}
// Traverse the suffix sum array
for(int i = N - 1; i >= 0; i--)
{
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - dp[i, K] >= K)
{
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 7, 4, 5, 6, 8 };
int K = 13;
int N = arr.Length;
Console.WriteLine(MinimumLength(arr, N, K));
}
}
// This code is contributed by ukasp
<script>
// javascript program for the above approach
var max1 = 1000000000;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
function MinimumLength(A, N, K)
{
0
// Sort the array in ascending order
A.sort();
// Stores suffix sum of the array
var suffix = Array(N + 1).fill(0);
var i;
// Update the suffix sum array
for (i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Stores all dp-states
var dp = new Array(N + 1);
for (i = 0; i < N+1; i++)
dp[i] = new Array(K + 1);
// Initialize all dp-states
// with a max1imum possible value
var j;
for (i = 0; i <= N; i++) {
for (j = 0; j <= K; j++){
dp[i][j] = max1;
}
};
// Base Case
dp[N][0] = 0;
// Traverse the array arr[]
for (i = N - 1; i >= 0; i--) {
// Iterate over the range [0, K]
for (j = K; j >= 0; j--) {
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i]) {
dp[i][j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (dp[i + 1][j - A[i]] == max1)
dp[i][j] = max1;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
dp[i][j] = Math.min(dp[i + 1][j],
dp[i + 1][j - A[i]] + A[i]);
}
}
// Traverse the suffix sum array
for (i = N - 1; i >= 0; i--) {
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - dp[i][K] >= K) {
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
var arr = [7, 4, 5, 6, 8];
var K = 13;
var N = arr.length;
document.write(MinimumLength(arr, N, K));
// This code is contributed by SURENDRA_GANGWAR.
</script>
Time Complexity: O(N * K)
Auxiliary Space: O(N * K)
Efficient Approach : using array instead of 2d matrix to optimize space complexity
In previous code we can se that dp[i][j] is dependent upon dp[i+1][j-1] or dp[i][j-1] so we can assume that dp[i+1] is next and dp[i] is current row.
Implementations Steps :
- Sort the array in ascending order and calculates the suffix sum of the array.
- Initializes two vectors curr and next with a maximum possible value, where curr represents the current state, and next represents the next state.
- Set the base case of curr and next vectors as 0 for the 0th index and traverses the array from N-1 index to 0.
- For each element A[i] in the array, it iterates over the range [0, K] in reverse order and updates the curr vector by choosing the minimum length of subsets with sum greater than or equal to K. It uses the next vector to calculate the minimum value.
- Now Updates the next vector as the curr vector.
- Finally traverses the suffix sum array in reverse order and finds the sum of lengths of two smallest subsets that have sum greater than or equal to K.
- Returns the sum of lengths if it is greater than or equal to K, otherwise, it returns -1.
Implementation :
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e9;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
int MinimumLength(int A[], int N, int K)
{
// Sort the array in ascending order
sort(A, A + N);
// Stores suffix sum of the array
int suffix[N + 1] = { 0 };
// Update the suffix sum array
for (int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Initialize all dp-states
// with a maximum possible value
vector<int>curr(K+1 , MAX);
vector<int>next(K+1 , MAX);
// Base Case
curr[0] = 0;
next[0] = 0;
// Traverse the array arr[]
for (int i = N - 1; i >= 0; i--) {
// Iterate over the range [0, K]
for (int j = K; j >= 0; j--) {
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i]) {
curr[j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (next[j - A[i]] == MAX)
curr[j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
curr[j] = min(next[j],
next[j - A[i]] + A[i]);
}
next = curr;
}
// Traverse the suffix sum array
for (int i = N - 1; i >= 0; i--) {
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - curr[K] >= K) {
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
int main()
{
int arr[] = { 7, 4, 5, 6, 8 };
int K = 13;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MinimumLength(arr, N, K);
return 0;
}
// this code is contributed by bhardwajji
import java.util.Arrays;
public class Main {
static final int MAX = (int) 1e9;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int[] A, int N, int K)
{
// Sort the array in ascending order
Arrays.sort(A);
// Stores suffix sum of the array
int[] suffix = new int[N + 1];
// Update the suffix sum array
for (int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Initialize all dp-states
// with a maximum possible value
int[] curr = new int[K + 1];
Arrays.fill(curr, MAX);
int[] next = new int[K + 1];
Arrays.fill(next, MAX);
// Base Case
curr[0] = 0;
next[0] = 0;
// Traverse the array arr[]
for (int i = N - 1; i >= 0; i--) {
// Iterate over the range [0, K]
for (int j = K; j >= 0; j--) {
if (j <= A[i]) {
curr[j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (next[j - A[i]] == MAX)
curr[j] = MAX;
else
curr[j] = Math.min(next[j],
next[j - A[i]] + A[i]);
}
next = curr.clone();
}
// Traverse the suffix sum array
for (int i = N - 1; i >= 0; i--) {
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - curr[K] >= K) {
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
public static void main(String[] args) {
int[] arr = {7, 4, 5, 6, 8};
int K = 13;
int N = arr.length;
System.out.println(MinimumLength(arr, N, K));
}
}
const MAX = 1e9;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
function MinimumLength(A, N, K) {
// Sort the array in ascending order
A.sort((a, b) => a - b);
// Stores suffix sum of the array
let suffix = Array(N + 1).fill(0);
// Update the suffix sum array
for (let i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Initialize all dp-states
// with a maximum possible value
let curr = Array(K + 1).fill(MAX);
let next = Array(K + 1).fill(MAX);
// Base Case
curr[0] = 0;
next[0] = 0;
// Traverse the array arr[]
for (let i = N - 1; i >= 0; i--) {
// Iterate over the range [0, K]
for (let j = K; j >= 0; j--) {
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i]) {
curr[j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (next[j - A[i]] == MAX)
curr[j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
curr[j] = Math.min(next[j],
next[j - A[i]] + A[i]);
}
next = [...curr];
}
// Traverse the suffix sum array
for (let i = N - 1; i >= 0; i--) {
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - curr[K] >= K) {
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
let arr = [7, 4, 5, 6, 8];
let K = 13;
let N = arr.length;
console.log(MinimumLength(arr, N, K));
import sys
MAX = sys.maxsize
# Function to calculate sum of lengths
# of two smallest subsets with sum >= K
def MinimumLength(A, N, K):
# Sort the array in ascending order
A.sort()
# Stores suffix sum of the array
suffix = [0] * (N+1)
# Update the suffix sum array
for i in range(N - 1, -1, -1):
suffix[i] = suffix[i + 1] + A[i]
# Initialize all dp-states
# with a maximum possible value
curr = [MAX] * (K+1)
next_ = [MAX] * (K+1)
# Base Case
curr[0] = 0
next_[0] = 0
# Traverse the array arr[]
for i in range(N - 1, -1, -1):
# Iterate over the range [0, K]
for j in range(K, -1, -1):
# If A[i] is equal to at
# least the required sum
# j for the current state
if j <= A[i]:
curr[j] = A[i]
continue
# If the next possible
# state doesn't exist
if next_[j - A[i]] == MAX:
curr[j] = MAX
# Otherwise, update the current
# state to the minimum of the
# next state and state including
# the current element A[i]
else:
curr[j] = min(next_[j],
next_[j - A[i]] + A[i])
next_ = curr.copy()
# Traverse the suffix sum array
for i in range(N - 1, -1, -1):
# If suffix[i] - dp[i][K] >= K
if suffix[i] - curr[K] >= K:
# Sum of lengths of the two
# smallest subsets is obtained
return N - i
# Return -1, if there doesn't
# exist any subset of sum >= K
return -1
# Driver Code
if __name__ == "__main__":
arr = [7, 4, 5, 6, 8]
K = 13
N = len(arr)
print(MinimumLength(arr, N, K))
using System;
using System.Collections.Generic;
using System.Linq;
public class Program
{
static int MAX = 1000000000;
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int[] A, int N, int K)
{
// Sort the array in ascending order
Array.Sort(A);
// Stores suffix sum of the array
int[] suffix = new int[N + 1];
// Update the suffix sum array
for (int i = N - 1; i >= 0; i--)
suffix[i] = suffix[i + 1] + A[i];
// Initialize all dp-states
// with a maximum possible value
List<int> curr = new List<int>(Enumerable.Repeat(MAX, K + 1));
List<int> next = new List<int>(Enumerable.Repeat(MAX, K + 1));
// Base Case
curr[0] = 0;
next[0] = 0;
// Traverse the array arr[]
for (int i = N - 1; i >= 0; i--)
{
// Iterate over the range [0, K]
for (int j = K; j >= 0; j--)
{
// If A[i] is equal to at
// least the required sum
// j for the current state
if (j <= A[i])
{
curr[j] = A[i];
continue;
}
// If the next possible
// state doesn't exist
if (next[j - A[i]] == MAX)
curr[j] = MAX;
// Otherwise, update the current
// state to the minimum of the
// next state and state including
// the current element A[i]
else
curr[j] = Math.Min(next[j], next[j - A[i]] + A[i]);
}
next = new List<int>(curr);
}
// Traverse the suffix sum array
for (int i = N - 1; i >= 0; i--)
{
// If suffix[i] - dp[i][K] >= K
if (suffix[i] - curr[K] >= K)
{
// Sum of lengths of the two
// smallest subsets is obtained
return N - i;
}
}
// Return -1, if there doesn't
// exist any subset of sum >= K
return -1;
}
// Driver Code
public static void Main()
{
int[] arr = { 7, 4, 5, 6, 8 };
int K = 13;
int N = arr.Length;
Console.WriteLine(MinimumLength(arr, N, K));
}
}
Time Complexity: O(N * K)
Auxiliary Space: O(K)
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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
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
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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
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
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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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