### nik09877's blog

By nik09877, history, 4 months ago, Came In the Recent Leetcode Contest,Don't know where my code went wrong Link To The Problem ~~~~~ You are given two integer arrays nums1 and nums2 of length n.

The XOR sum of the two integer arrays is (nums1 XOR nums2) + (nums1 XOR nums2) + ... + (nums1[n — 1] XOR nums2[n — 1]) (0-indexed).

For example, the XOR sum of [1,2,3] and [3,2,1] is equal to (1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4. Rearrange the elements of nums2 such that the resulting XOR sum is minimized.

Return the XOR sum after the rearrangement. ~~~~~

Constraints:

n == nums1.length
n == nums2.length
1 <= n <= 14
0 <= nums1[i], nums2[i] <= 10^7

int dp[(1 << 21)];
class Solution
{
public:
int solve(int i, int mask, int &n, vector<int> &a, vector<int> &b)
{
if (i == n)
return 0;

for (int j = 0; j < n; j++)
{
if (mask & (1 << j))
answer = min(answer, a[i] ^ b[j] + solve(i + 1, (mask ^ (1 << j)), n, a, b));
}

}
int minimumXORSum(vector<int> &a, vector<int> &b)
{
memset(dp, -1, sizeof dp);
int n = a.size();
return solve(0, (1 << n) - 1, n, a, b);
}
};


By nik09877, history, 8 months ago, Getting WA on test case 4 . The link to the question https://codeforces.com/edu/course/2/lesson/6/3/practice/contest/285083/problem/B Can anyone help? what am i doing wrong?

#include <bits/stdc++.h>
#include <stdio.h>
#define repll(i, a, n) for (lli i = a; i <= n; i++)
#define lli long long int
#define ii int, int
#define pii pair<int, int>
#define pll pair<long, long>
#define plli pair<long long int, long long int>
#define vi vector<int>
#define vvi vector<vector<int>>
#define vlli vector<long long int>
#define endl '\n'
#define fastio ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0)
using namespace std;
const int mod = 1000000007;

lli n, k, l, r, mid, ans(-1);
vlli a;

lli segments(lli target)
{
lli cnt(0), sum(0);
repll(i, 0, n - 1)
{
if (a[i] > target)
return k + 1;
sum += a[i];
if (sum > target)
{
cnt++;
sum = a[i];
}
}
cnt++;
return cnt;
}
void solve()
{
cin >> n >> k;
a.resize(n);

repll(i, 0, n - 1) cin >> a[i], r += a[i];

l = 0, r += 10000;

while (l <= r)
{
mid = (l + r) / 2;
lli seg = segments(mid);
if (seg == k)
{
ans = mid;
r = mid - 1;
}
else if (seg > k)
{
l = mid + 1;
}
else
r = mid - 1;
}

cout << ans << endl;
return;
}
int32_t main()
{
fastio;
int t = 1;
while (t--)
{
solve();
}
}