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Commit 35bb941

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Merge pull request #102 from jineshparakh/master
Added Dynamic Programming and Recursion Problems
2 parents 5a93c09 + 6bd698b commit 35bb941

10 files changed

+517
-0
lines changed

‎10- Recursion & Backtracking/LCS.cpp

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#include<bits/stdc++.h>
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using namespace std;
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#define ll long long
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ll LCS(string X, string Y, ll n, ll m) {
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/*
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Base case:
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If either one of the two strings becomes null string, we cannot choose
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more from it.
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So return 0.
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*/
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if (n == 0 || m == 0)
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return 0LL;
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/*
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Choice Diagram:
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A. If the n-1th and m-1th character of both strings is equal, we select it and recur
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for n-1 and m-1
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B. If they are not equal, we have two choices.
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1. Recur for n-1,m
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2. Recur for n,m-1
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We have to find the LCS, so we will choose the max among the two cases.
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*/
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if (X[n - 1] == Y[m - 1])
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return LCS(X, Y, n - 1, m - 1) + 1LL;
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else {
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return max(LCS(X, Y, n - 1, m), LCS(X, Y, n, m - 1));
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}
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}
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int main() {
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/*
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LONGEST COMMON SUBSEQUENCE LENGTH
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Given two string X and Y, find the length of LCS between them.
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*/
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string X, Y; cin >> X >> Y;
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ll n = X.size(), m = Y.size();
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cout << LCS(X, Y, n, m) << endl;
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/*
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Sample Input:
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AGGTAB
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GXTXAYB
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Output:
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4
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Sample Input:
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ABCDGH
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AEDFHR
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Output:
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3
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*/
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return 0;
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}
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#include<bits/stdc++.h>
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using namespace std;
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#define ll long long
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ll findCountofSubsets(ll a[], ll n, ll S) {
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if (n == 0 && S == 0)
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return 1;
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if (n == 0 && S != 0)
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return 0;
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if (n != 0 && S == 0)
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return 1;
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if (a[n - 1] <= S) {
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return findCountofSubsets(a, n - 1, S) + findCountofSubsets(a, n - 1, S - a[n - 1]);
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}
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else return findCountofSubsets(a, n - 1, S);
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}
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int main() {
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/*
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Given n as the size of array
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a[i] is the ith element of the array
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S is the requeired absolute difference between the two partitions of the array
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To find:
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Count the number of valid pairs of subsets such that the given difference
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is obtained
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*/
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ll n, S; cin >> n >> S;
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ll a[n];
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ll sum = 0;
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for(ll i = 0; i < n; i++) {
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cin >> a[i];
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sum += a[i];
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}
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if ((sum + S) & 1)
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cout << 0 << endl;
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else
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/*
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Why (sum+S)/2;
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sum(subset1)-sum(subset2)=difference(S);
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sum(subset1)+sub(subset2)=sumofArray(sum);
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Add these two
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sum(subset1)=(S+sum)/2
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*/
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cout << findCountofSubsets(a, n, (sum + S) / 2) << endl;
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return 0;
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}
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#include <bits/stdc++.h>
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using namespace std;
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#define ll long long
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ll countSubsets(ll a[], ll n, ll S) {
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if (n == 0 && S == 0)
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return 1;
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else if (n != 0 && S == 0)
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return 1;
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else if (n == 0 && S != 0)
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return 0;
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if (a[n - 1] <= S) {
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return countSubsets(a, n - 1, S) + countSubsets(a, n - 1, S - a[n - 1]);
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}
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else
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return countSubsets(a, n - 1, S);
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}
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int main() {
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/*
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Given n as the size of the array and S as the required sum.
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a[i] is the ith element of the array which is >0
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To find:
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The count of subsets having the sum S
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Similar to the subset sum problem, just instead of || use +
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*/
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ll n, S;
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cin >> n >> S;
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ll a[n];
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for (ll i = 0; i < n; i++)
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cin >> a[i];
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cout << countSubsets(a, n, S) << endl;
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/*
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INPUT:
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6 10
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2 3 5 6 8 10
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OUTPUT:
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3
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*/
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return 0;
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}
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#include<bits/stdc++.h>
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using namespace std;
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#define ll long long
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void deleteMiddleElement(stack<ll> &s, ll n) {
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if (s.empty())
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return;
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ll val = s.top();
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s.pop();
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deleteMiddleElement(s, n - 1);
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if (n != 0) // if the element is middle element then do not add it
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s.push(val);
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}
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int main() {
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stack<ll> s;
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ll n; cin >> n;
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for (ll i = 0; i < n; i++) {
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ll x; cin >> x;
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s.push(x);
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}
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deleteMiddleElement(s, s.size() / 2);
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while (!s.empty()) {
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cout << s.top() << " ";
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s.pop();
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}
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}
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#include<bits/stdc++.h>
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using namespace std;
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#define ll long long
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ll editDistance(string X, string Y, ll n, ll m) {
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/*
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The base conditions:
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If either of the strings is empty you need the corresponding lengths to
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make them same
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If the characters are same do nothing and go for a smaller input
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If not, we need to do something which costs 1 and them we choose the least cost amongst
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insert, replace and remove.
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Replace-> call for(n-1,m-1)
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Insert-> call for(n,m-1)
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Remove->call for(n-1,m)
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*/
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if (n == 0)
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return m;
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if (m == 0)
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return n;
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if (X[n - 1] == Y[m - 1])
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return editDistance(X, Y, n - 1, m - 1);
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else
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return 1 + min(editDistance(X, Y, n - 1, m - 1), min(editDistance(X, Y, n - 1, m), editDistance(X, Y, n, m - 1)));
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}
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int main() {
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/*
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Edit Distance.
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Convert String X to string Y by doing insert, replace and remove.
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All three are of the same cost.
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Find the minimum cost of conversion
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*/
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string X, Y; cin >> X >> Y;
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cout << editDistance(X, Y, X.size(), Y.size());
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return 0;
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}
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#include <bits/stdc++.h>
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using namespace std;
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#define ll long long
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bool equalSumPartition(ll a[], ll n, ll S) {
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if (n == 0 && S == 0)
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return 1;
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if (n == 0 && S != 0)
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return 0;
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if (n != 0 && S == 0)
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return 1;
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if (a[n - 1] <= S) {
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return equalSumPartition(a, n - 1, S) || equalSumPartition(a, n - 1, S - a[n - 1]);
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}
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else
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return equalSumPartition(a, n - 1, S);
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}
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int main() {
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/*
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Given Input:
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n denotes the number of elements in the array
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a[i] is the ith element of the array
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S is the sum of the array
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To find:
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Return True if the array can be divided into 2 subsets of equal sum else return false
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Interpretations:
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Problem is similar to 0-1 knapsack.
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We gotta make choices.
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a. include it
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b. don't include the element.
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*/
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ll n, S=0;
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cin >> n;
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ll a[n];
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for (ll i = 0; i < n; i++)
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cin >> a[i], S += a[i];
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if (S & 1)
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cout << "NO" << endl;
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else
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cout << (equalSumPartition(a, n, S / 2) ? "YES" : "NO") << endl;
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/*
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Sample Input:
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5
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2 3 5 7 8
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Output:
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NO
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Sample Input:
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5
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5 1 1 1 2
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Output: YES
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*/
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return 0;
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}
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#include <bits/stdc++.h>
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using namespace std;
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#define ll long long
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bool subsetSum(ll a[], ll n, ll S)
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{
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/*
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Base case:
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If n==0 && S==0 we can have an empty subset
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If n==0 && S!=0 we cannot have empty subset
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If n!=0 && S==0 we can choose an empty subset
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Think of the smallest possible input which keeps on changing in the
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revursive code.
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*/
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if (n == 0 && S == 0)
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return 1;
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else if (n == 0 && S != 0)
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return 0;
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else if (n != 0 && S == 0)
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return 1;
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/*
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The choice diagram.
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Start from the back.
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We used max in the 0-1 knapsack, here we use || because if either of the two sums
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is possible we should get a 1. Similar to the boolean max thing(is || only).
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*/
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if (a[n - 1] <= S) {
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return subsetSum(a, n - 1, S - a[n - 1]) || subsetSum(a, n - 1, S);
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}
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else {
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return subsetSum(a, n - 1, S);
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}
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}
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int main() {
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/*
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Given Input:
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n denotes the number of elements in the array
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a[i] is the ith element of the array
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S is the target sum.
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To find:
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Return True if any subset with the given sum exists else return false.
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Interpretations:
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Problem is similar to 0-1 knapsack.
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We gotta make choices.
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a. include it
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b. don't include the element.
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*/
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ll n, S;
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cin >> n >> S;
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ll a[n];
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for (ll i = 0; i < n; i++)
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cin >> a[i];
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cout << (subsetSum(a, n, S) ? "YES" : "NO") << endl;
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/*
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Sample Input:
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5 1
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2 3 5 7 8
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Output:
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NO
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Sample Input:
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5 11
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2 3 5 7 8
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Output: YES
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*/
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return 0;
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}

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