Find the node whose sum with X has minimum set bits

Given a tree, and the weights of all the nodes and an integer x, the task is to find a node i such that weight[i] + x gives the minimum setbits, If two or more nodes have the same count of set bits when added with x then find the one with the minimum value.



x = 15
Output: 1
Node 1: setbits(5 + 15) = 2
Node 2: setbits(10 + 15) = 3
Node 3: setbits(11 + 15) = 3
Node 4: setbits(8 + 15) = 4
Node 5: setbits(6 + 15) = 3

Approach: Perform dfs on the tree and keep track of the node whose sum with x has minimum set bits. If two or more nodes have equal count of set bits then choose the one with the minimum number.

Below is the implementation of the above approach:





// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int minimum = INT_MAX, x, ans = INT_MAX;
vector<int> graph[100];
vector<int> weight(100);
// Function to perform dfs to find
// the minimum set bits value
void dfs(int node, int parent)
    // If current set bits value is smaller than
    // the current minimum
    int a = __builtin_popcount(weight[node] + x);
    if (minimum > a) {
        minimum = a;
        ans = node;
    // If count is equal to the minimum
    // then choose the node with minimum value
    else if (minimum == a)
        ans = min(ans, node);
    for (int to : graph[node]) {
        if (to == parent)
        dfs(to, node);
// Driver code
int main()
    x = 15;
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
    // Edges of the tree
    dfs(1, 1);
    cout << ans;
    return 0;




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