Given an array a of size N. The task is to find a subsequence X of length K such that gcd(X[0], X[1]) + (X[2], X[3]) + … is maximum.
Note: K is even.
Examples:
Input: a[] = {4, 5, 3, 7, 8, 10, 9, 8}, k = 4
Output: 9
The subsequence {4, 7, 8, 8} gives the maximum value = 9.
Other subsequences like { 5, 3, 8, 8} also gives 9.
Input: a[] = {5, 8, 16, 20}, K = 2
Output:
The subsequence {8, 16} gives the maximum value.
Naive Approach: A naive approach is to generate all subsequences of length K using recursion and find the maximum value possible.
Efficient Approach: An efficient approach is to use Dynamic Programming to solve the above problem. Let dp[i][j] denote the value of the pair gcd sum if i-th index element is taken as the j-th element. Recursively generate all subsequences possible. If cnt of elements taken is odd then add gcd(a[prev], a[current]) to the sum since this number will be the second in the pair and recur for next element. If cnt is even, then simply recur setting a[i] as the previous element. Memoization has been used to avoid re-calculation of the same recurrence. The maximum of all the generated sums will be the answer.
Below is the implementation of the above approach:
C++
// C++ program to find the sum of// the addition of all possible subsets#include <bits/stdc++.h>using namespace std;const int MAX = 100;// Recursive function to find the maximum value of// the given recurrenceint recur(int ind, int cnt, int last, int a[], int n, int k, int dp[][MAX]){ // If we get K elements if (cnt == k) return 0; // If we have reached the end // and K elements are not there if (ind == n) return -1e9; // If the state has been visited if (dp[ind][cnt] != -1) return dp[ind][cnt]; int ans = 0; // Iterate for every element as the // next possible element // and take the element which gives // the maximum answer for (int i = ind; i < n; i++) { // If this element is the first element // in the individual pair in the subsequence // then simply recurrence with the last // element as i-th index if (cnt % 2 == 0) ans=max(ans,recur(i + 1, cnt + 1, i, a, n, k, dp)); // If this element is the second element in // the individual pair, the find gcd with // the previous element and add to the answer // and recur for the next element else ans = max(ans, __gcd(a[last], a[i]) + recur(i + 1, cnt + 1, 0, a, n, k, dp)); } return dp[ind][cnt] = ans;}// Driver Codeint main(){ int a[] = { 4, 5, 3, 7, 8, 10, 9, 8 }; int n = sizeof(a) / sizeof(a[0]); int k = 4; int dp[n][MAX]; memset(dp, -1, sizeof dp); cout << recur(0, 0, 0, a, n, k, dp);} |
Java
// Java program to find the sum of// the addition of all possible subsetsimport java.util.*;class GFG {static int MAX = 100;// Recursive function to find the maximum value of// the given recurrencestatic int recur(int ind, int cnt, int last, int a[], int n, int k, int dp[][]){ // If we get K elements if (cnt == k) return 0; // If we have reached the end // and K elements are not there if (ind == n) return (int) -1e9; // If the state has been visited if (dp[ind][cnt] != -1) return dp[ind][cnt]; int ans = 0; // Iterate for every element as the // next possible element // and take the element which gives // the maximum answer for (int i = ind; i < n; i++) { // If this element is the first element // in the individual pair in the subsequence // then simply recurrence with the last // element as i-th index if (cnt % 2 == 0) ans = Math.max(ans,recur(i + 1, cnt + 1, i, a, n, k, dp)); // If this element is the second element in // the individual pair, the find gcd with // the previous element and add to the answer // and recur for the next element else ans = Math.max(ans, __gcd(a[last], a[i]) + recur(i + 1, cnt + 1, 0, a, n, k, dp)); } return dp[ind][cnt] = ans;}static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); }// Driver Codepublic static void main(String[] args){ int a[] = { 4, 5, 3, 7, 8, 10, 9, 8 }; int n = a.length; int k = 4; int [][]dp = new int[n][MAX]; for(int i = 0; i < n; i++) { for(int j = 0; j < MAX; j++) { dp[i][j] = -1; } } System.out.println(recur(0, 0, 0, a, n, k, dp)); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find the sum of# the addition of all possible subsetsfrom math import gcd as __gcdMAX = 100# Recursive function to find the # maximum value of the given recurrencedef recur(ind, cnt, last, a, n, k, dp): # If we get K elements if (cnt == k): return 0 # If we have reached the end # and K elements are not there if (ind == n): return -10**9 # If the state has been visited if (dp[ind][cnt] != -1): return dp[ind][cnt] ans = 0 # Iterate for every element as the # next possible element # and take the element which gives # the maximum answer for i in range(ind, n): # If this element is the first element # in the individual pair in the subsequence # then simply recurrence with the last # element as i-th index if (cnt % 2 == 0): ans = max(ans, recur(i + 1, cnt + 1, i, a, n, k, dp)) # If this element is the second element in # the individual pair, the find gcd with # the previous element and add to the answer # and recur for the next element else: ans = max(ans, __gcd(a[last], a[i]) + recur(i + 1, cnt + 1, 0, a, n, k, dp)) dp[ind][cnt] = ans return ans# Driver Codea = [4, 5, 3, 7, 8, 10, 9, 8 ]n = len(a)k = 4;dp = [[-1 for i in range(MAX)] for i in range(n)]print(recur(0, 0, 0, a, n, k, dp))# This code is contributed by Mohit Kumar |
C#
// C# program to find the sum of// the addition of all possible subsetsusing System; class GFG {static int MAX = 100;// Recursive function to find the maximum value of// the given recurrencestatic int recur(int ind, int cnt, int last, int []a, int n, int k, int [,]dp){ // If we get K elements if (cnt == k) return 0; // If we have reached the end // and K elements are not there if (ind == n) return (int) -1e9; // If the state has been visited if (dp[ind,cnt] != -1) return dp[ind,cnt]; int ans = 0; // Iterate for every element as the // next possible element // and take the element which gives // the maximum answer for (int i = ind; i < n; i++) { // If this element is the first element // in the individual pair in the subsequence // then simply recurrence with the last // element as i-th index if (cnt % 2 == 0) ans = Math.Max(ans,recur(i + 1, cnt + 1, i, a, n, k, dp)); // If this element is the second element in // the individual pair, the find gcd with // the previous element and add to the answer // and recur for the next element else ans = Math.Max(ans, __gcd(a[last], a[i]) + recur(i + 1, cnt + 1, 0, a, n, k, dp)); } return dp[ind,cnt] = ans;}static int __gcd(int a, int b) { if (b == 0) return a; return __gcd(b, a % b); }// Driver Codepublic static void Main(String[] args){ int []a = { 4, 5, 3, 7, 8, 10, 9, 8 }; int n = a.Length; int k = 4; int [,]dp = new int[n,MAX]; for(int i = 0; i < n; i++) { for(int j = 0; j < MAX; j++) { dp[i,j] = -1; } } Console.WriteLine(recur(0, 0, 0, a, n, k, dp));}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program to find the sum of// the addition of all possible subsetslet MAX = 100;// Recursive function to find the maximum value of// the given recurrencefunction recur(ind,cnt,last,a,n,k,dp){ // If we get K elements if (cnt == k) return 0; // If we have reached the end // and K elements are not there if (ind == n) return -1e9; // If the state has been visited if (dp[ind][cnt] != -1) return dp[ind][cnt]; let ans = 0; // Iterate for every element as the // next possible element // and take the element which gives // the maximum answer for (let i = ind; i < n; i++) { // If this element is the first element // in the individual pair in the subsequence // then simply recurrence with the last // element as i-th index if (cnt % 2 == 0) ans = Math.max(ans,recur(i + 1, cnt + 1, i, a, n, k, dp)); // If this element is the second element in // the individual pair, the find gcd with // the previous element and add to the answer // and recur for the next element else ans = Math.max(ans, __gcd(a[last], a[i]) + recur(i + 1, cnt + 1, 0, a, n, k, dp)); } return dp[ind][cnt] = ans;}function __gcd(a,b){ if (b == 0) return a; return __gcd(b, a % b);}// Driver Codelet a=[4, 5, 3, 7, 8, 10, 9, 8];let n = a.length;let k = 4;let dp = new Array(n);for(let i=0;i<n;i++){ dp[i]=new Array(MAX); for(let j = 0; j < MAX; j++) { dp[i][j] = -1; }}document.write(recur(0, 0, 0, a, n, k, dp));// This code is contributed by unknown2108</script> |
Output:
9
Time Complexity: O(N*k), as we are using a loop to traverse N times and in each traversal we are recursively calling the function which will cost O(k) time. Where N is the number of elements in the array.
Auxiliary Space: O(N*100), as we are using extra space for the dp matrix. Where N is the number of elements in the array.
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