Given an array A of n integers and given queries in the form of range [l , r] and a value x, find the minimum of A[i] XOR x where l <= i <= r and x will be different for different queries.
I tried solving this problem using segment trees but I am not sure what type of information I should store in them as x will be different for different queries.
0 < number of queries <= 1e4
0 < n <= 1e4
To solve this I used a std::vector as basis (not an array, or std::array), just for flexibility.
#include <algorithm>
#include <stdexcept>
#include <vector>
int get_xored_max(const std::vector<int>& values, const size_t l, const size_t r, const int xor_value)
{
// check bounds of l and r
if ((l >= values.size()) || (r >= values.size()))
{
throw std::invalid_argument("index out of bounds");
}
// todo check l < r
// create left & right iterators to create a smaller vector
// only containing the subset we're interested in.
auto left = values.begin() + l;
auto right = values.begin() + r + 1;
std::vector<int> range{ left, right };
// xor all the values in the subset
for (auto& v : range)
{
v ^= xor_value;
}
// use the standard library function for finding the iterator to the maximum
// then use the * to dereference the iterator and get the value
auto max_value = *std::max_element(range.begin(), range.end());
return max_value;
}
int main()
{
std::vector<int> values{ 1,3,5,4,2,4,7,9 };
auto max_value = get_xored_max(values, 0u, 7u, 3);
return 0;
}
Approach - Trie + Offline Processing
Time Complexity - O(N32)
Space Complexity - O(N32)
Edit:
This Approach will fail. I guess, we have to use square root decomposition instead of two pointers approach.
I have solved this problem using Trie for finding minimum xor in a range of [l,r]. I solved queries by offline processing by sorting them.
Input format:
the first line has n (no. of elements) and q (no. of queries). the second line has all n elements of the array. each subsequent line has a query and each query has 3 inputs l, r and x.
Example -
Input -
3 3
2 1 2
1 2 3
1 3 2
2 3 5
First, convert all 3 queries into queries sorted by l and r.
converted queries -
1 2 3
1 3 2
2 3 5
Key here is processing over sorted queries using two pointers approach.
#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e4 + 77;
int n, q, l, r, x;
int a[N], ans[N];
vector<pair<pair<int, int>, pair<int, int>>> queries;
// Trie Implementation starts
struct node
{
int nxt[2], cnt;
void newnode()
{
memset(nxt, 0, sizeof(nxt));
cnt = 0;
}
} trie[N * 32];
int tot = 1;
void update(int x, int v)
{
int p = 1;
for (int i = 31; i >= 0; i--)
{
int id = x >> i & 1;
if (!trie[p].nxt[id])
{
trie[++tot].newnode();
trie[p].nxt[id] = tot;
}
p = trie[p].nxt[id];
trie[p].cnt += v;
}
}
int minXor(int x)
{
int res = 0, p = 1;
for (int i = 31; i >= 0; i--)
{
int id = x >> i & 1;
if (trie[p].nxt[id] and trie[trie[p].nxt[id]].cnt)
p = trie[p].nxt[id];
else
{
p = trie[p].nxt[id ^ 1];
res |= 1 << i;
}
}
return res;
}
// Trie Implementation ends
int main()
{
cin >> n >> q;
for (int i = 1; i <= n; i += 1)
{
cin >> a[i];
}
for (int i = 1; i <= q; i += 1)
{
cin >> l >> r >> x;
queries.push_back({{l, r}, {x, i}});
}
sort(queries.begin(), queries.end());
int left = 1, right = 1;
for (int i = 0; i < q; i += 1)
{
int l = queries[i].first.first;
int r = queries[i].first.second;
int x = queries[i].second.first;
int index = queries[i].second.second;
while (left < l)
{
update(a[left], -1);
left += 1;
}
while (right <= r)
{
update(a[right], 1);
right += 1;
}
ans[index] = minXor(x);
}
for (int i = 1; i <= q; i += 1)
{
cout << ans[i] << " \n";
}
return 0;
}
Edit: with O(number of bits) code
Use a binary tree to store the values of A, look here : Minimum XOR for queries
What you need to change is adding to each node the range of indexes for A corresponding to the values in the leafs.
# minimal xor in a range
nbits=16 # Number of bits for numbers
asize=5000 # Array size
ntest=50 # Number of random test
from random import randrange
# Insert element a iindex iin the tree (increasing i only)
def tinsert(a,i,T):
for b in range(nbits-1,-1,-1):
v=((a>>b)&1)
T[v+2].append(i)
if T[v]==[]:T[v]=[[],[],[],[]]
T=T[v]
# Buildtree : builds a tree based on array V
def build(V):
T=[[],[],[],[]] # Init tree
for i,a in enumerate(V): tinsert(a,i,T)
return(T)
# Binary search : is T intersec [a,b] non empty ?
def binfind(T,a,b):
s,e,om=0,len(T)-1,-1
while True:
m=(s+e)>>1
v=T[m]
if v<a:
s=m
if m==om: return(a<=T[e]<=b)
elif v>b:
e=m
if m==om: return(a<=T[s]<=b)
else: return(True) # a<=T(m)<=b
om=m
# Look for the min xor in a give range index
def minx(x,s,e,T):
if s<0 or s>=(len(T[2])+len(T[3])) or e<s: return
r=0
for b in range(nbits-1,-1,-1):
v=((x>>b)&1)
if T[v+2]==[] or not binfind(T[v+2],s,e): # not nr with b set to v ?
v=1-v
T=T[v]
r=(r<<1)|v
return(r)
# Tests the code on random arrays
max=(1<<nbits)-1
for i in range(ntest):
A=[randrange(0,max) for i in range(asize)]
T=build(A)
x,s=randrange(0,max),randrange(0,asize-1)
e=randrange(s,asize)
if min(v^x for v in A[s:e+1])!=x^minx(x,s,e,T):
print('error')
I was able to solve this using segment tree and tries as suggested by @David Eisenstat
Below is an implementation in c++. I constructed a trie for each segment in the segment tree. And finding the minimum xor is just traversing and matching the corresponding trie using each bit of the query value (here)
#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i < b; i++)
using namespace std;
const int bits = 7;
struct trie {
trie *children[2];
bool end;
};
trie *getNode(void)
{
trie *node = new trie();
node->end = false;
node->children[0] = NULL;
node->children[1] = NULL;
return node;
}
trie *merge(trie *l, trie *r)
{
trie *node = getNode();
// Binary 0:
if (l->children[0] && r->children[0])
node->children[0] = merge(l->children[0], r->children[0]);
else if (!r->children[0])
node->children[0] = l->children[0];
else if (!l->children[0])
node->children[0] = r->children[0];
// Binary 1:
if (l->children[1] && r->children[1])
node->children[1] = merge(l->children[1], r->children[1]);
else if (!r->children[1])
node->children[1] = l->children[1];
else if (!l->children[1])
node->children[1] = r->children[1];
return node;
}
void insert(trie *root, int num)
{
int mask = 1 << bits;
int bin;
rep(i, 0, bits + 1)
{
bin = ((num & mask) >> (bits - i));
if (!root->children[bin]) root->children[bin] = getNode();
root = root->children[bin];
mask = mask >> 1;
}
root->end = true;
}
struct _segTree {
int n, height, size;
vector<trie *> tree;
_segTree(int _n)
{
n = _n;
height = (int)ceil(log2(n));
size = (int)(2 * pow(2, height) - 1);
tree.resize(size);
}
trie *construct(vector<int> A, int start, int end, int idx)
{
if (start == end) {
tree[idx] = getNode();
insert(tree[idx], A[start]);
return tree[idx];
}
int mid = start + (end - start) / 2;
tree[idx] = merge(construct(A, start, mid, 2 * idx + 1),
construct(A, mid + 1, end, 2 * idx + 2));
return tree[idx];
}
int findMin(int num, trie *root)
{
int mask = 1 << bits;
int bin;
int rnum = 0;
int res = 0;
rep(i, 0, bits + 1)
{
bin = ((num & mask) >> (bits - i));
if (!root->children[bin]) {
bin = 1 - bin;
if (!root->children[bin]) return res ^ num;
}
rnum |= (bin << (bits - i));
root = root->children[bin];
if (root->end) res = rnum;
mask = mask >> 1;
}
return res ^ num;
}
int Query(int X, int start, int end, int qstart, int qend, int idx)
{
if (qstart <= start && qend >= end) return findMin(X, tree[idx]);
if (qstart > end || qend < start) return INT_MAX;
int mid = start + (end - start) / 2;
return min(Query(X, start, mid, qstart, qend, 2 * idx + 1),
Query(X, mid + 1, end, qstart, qend, 2 * idx + 2));
}
};
int main()
{
int n, q;
vector<int> A;
vector<int> L;
vector<int> R;
vector<int> X;
cin >> n;
A.resize(n, 0);
rep(i, 0, n) cin >> A[i];
cin >> q;
L.resize(q);
R.resize(q);
X.resize(q);
rep(i, 0, q) cin >> L[i] >> R[i] >> X[i];
//---------------------code--------------------//
_segTree segTree(n);
segTree.construct(A, 0, n - 1, 0);
rep(i, 0, q)
{
cout << segTree.Query(X[i], 0, n - 1, L[i], R[i], 0) << " ";
}
return 0;
}
Time complexity : O((2n - 1)*k + qklogn)
Space complexity : O((2n - 1)*2k)
k -> number of bits