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Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.

Subset Sum Problem | DP-25 Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum. Example: Input:  set[] = {3, 34, 4, 12, 5, 2}, sum = 9 Output:  True  //There is a subset (4, 5) with sum 9. // A recursive solution for subset sum problem #include <stdio.h>     // Returns true if there is a subset of set[] with sun equal to given sum bool isSubsetSum( int set[], int n, int sum) {     // Base Cases     if (sum == 0)       return true ;     if (n == 0 && sum != 0)       return false ;         // If last element is greater than sum, then ignore it     if (set[n-1] > sum)       return isSubsetSum(set, n-1, sum);         /* else, check if sum...
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Print all subsets of given size of a set

Generate all possible subset of size  r  of given array with distinct elements. Examples: Input : arr[] = {1, 2, 3, 4} r = 2 Output : 1 2 1 3 1 4 2 3 2 4 3 4 Input : arr[] = {10, 20, 30, 40, 50} r = 3 Output : 10 20 30 10 20 40 10 20 50 10 30 40 10 30 50 10 40 50 20 30 40 20 30 50 20 40 50 30 40 50 The idea here is similar to  Subset Sum Problem . We one by one consider every element of input array, and recur for two cases: 1) The element is included in current combination (We put the element in data[] and increment next available index in data[]) 2) The element is excluded in current combination (We do not put the element and do not change index) When number of elements in data[] become equal to r (size of a combination), we print it. This method is mainly based on  Pascal’s Identity , i...
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[17 Feb 2020] Given an array of integers, find the nearest smaller number for every element such that the smaller element is on left side.

Find the nearest smaller numbers on left side in an array Given an array of integers, find the nearest smaller number for every element such that the smaller element is on left side. Examples: Input:  arr[] = {1, 6, 4, 10, 2, 5} Output:         {_, 1, 1,  4, 1, 2} First element ('1') has no element on left side. For 6, there is only one smaller element on left side '1'. For 10, there are three smaller elements on left side (1, 6 and 4), nearest among the three elements is 4. Input: arr[] = {1, 3, 0, 2, 5} Output:        {_, 1, _, 0, 2} Expected time complexity is O(n). // C++ implementation of efficient algorithm to find // smaller element on left side #include <iostream> #include <stack> using namespace std;   // Prints smaller elements on left side of every element void printPrevSmaller(int arr[], int n) {     // Create an empty stack     stack<int> S;   ...
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