HackerEarth Shubham and Subarray Xor problem solution

YASH PAL, 31 July 202414 February 2026

In this HackerEarth Shubham and Subarray Xor problem solution You are given an array consisting of n integers a1,a2,…an. Find the maximum value of xor of sum of 2 disjoint subarrays i.e maximize ( sum(l1,r1) xor sum(l2,r2) )

where 1 <= l1 <= r1 < l2 <= r2 <= n.

Note: sum(l,r) denotes the sum of elements from indices l to r both inclusive.

HackerEarth Shubham and Subarray Xor problem solution

HackerEarth Shubham and Subarray Xor problem solution.

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define inf 1000000000000
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define S second
#define F first
#define boost1 ios::sync_with_stdio(false);
#define boost2 cin.tie(0);
#define mem(a,val) memset(a,val,sizeof a)
#define endl "n"
#define maxn 820000

ll tr[8*maxn][2],nn=1,cnt[8*maxn][2],arr[905],sum[905];
void add(ll a) {
    ll t = 1;
    for (ll i = 31; i >= 0; --i) {
        int now = (a >> i) & 1;
        if (!tr[t][now]) {
            tr[t][now] = ++nn;
        }
        cnt[t][now]++;
        t = tr[t][now];
    }
}
ll getMax(ll a) {
    ll t = 1, res = 0;
    for (ll i = 31; i >= 0; --i) {
        ll now = (a >> i) & 1;
        now = !now;
        if (tr[t][now] && cnt[t][now]) {
            t = tr[t][now];
            res += (1 << i) * 1;
        } else {
            t = tr[t][!now];
        }
    }
    return res;
}
inline ll getsum(ll l,ll r)
{
    assert(l<=r);
    return sum[r]-sum[l-1];
}
int main()
{
    boost1;boost2;
    ll i,j,n,x,y,val,ans;
    cin>>n;
    for(i=1;i<=n;i++)
    cin>>arr[i];
    sum[1]=arr[1];
    for(i=2;i<=n;i++)
    sum[i]=sum[i-1]+arr[i];
    add(arr[n]);
    ans=0;
    for(i=n-1;i>=1;i--)
    {
        for(j=i;j>=1;j--)
        {
            val=getsum(j,i);
            ans=max(ans,getMax(val));
        }
        for(j=i;j<=n;j++)
        {   
            val=getsum(i,j);
            add(val);
        }
    }
    cout<<ans;

    return 0;
}

Second solution

#include <bits/stdc++.h>
using namespace std;

#define next _nxt

const int N = 10000005;
int sz = 0, next[2][N], arr[905], sum[905];
bool created[N];

void insert (int s) {
    int v = 0;
    for (int i = 30; i >= 0; i--) {
        int c = (s >> i) & 1;
        if (!created[next[c][v]]) {
            next[c][v] = ++sz;
            created[sz] = true;
        }
        v = next[c][v];
    }
}

int search (int tmp) {
    int v = 0, ans = 0;
    for (int i = 30; i >= 0; i--) {
        int c = (tmp >> i) & 1;
        if(created[next[1 ^ c][v]]){
            ans |= ((1 ^ c) << i);
            v = next[1 ^ c][v];
        }
        else{
            ans |= (c << i);
            v = next[c][v];
        }
    }
    return ans;
}

int main(){
    int i,j,n,maxi = 0,curr;
    scanf("%d", &n);
    for(i = 1; i <= n; i++){
        scanf("%d", &arr[i]);
        sum[i] = sum[i - 1] + arr[i];
    }
    for(i = 1; i <= n; i++){
        for(j = 1; j <= i; j++)
            insert(sum[i] - sum[j - 1]);
        for(j = i + 1; j <= n; j++)
            maxi = max(maxi, (sum[j] - sum[i]) ^ search(sum[j] - sum[i]));
    }
    printf("%dn", maxi);
    return 0;
}

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