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XOR Subsequences – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Solutions of Algorithms Data Structures Hard HackerRank:
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C++ replace HackerRank Solution
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <string>
using namespace std;
const int MAXN = 100005, MAX = (1 << 16);
int mem[MAXN], N, sum[MAXN];
long long total[1 << 16];
inline void walk(int digits) {
int n = (1 << digits);
for(int i = 1 ; i <= digits ; i++) {
int m = (1 << i);
int mh = m >> 1;
for(int r = 0 ; r < n ; r += m) {
int t1 = r;
int t2 = r + mh;
for(int j = 0 ; j < mh ; j++, t1++, t2++) {
long long u = total[t1];
long long v = total[t2];
total[t1] = u + v;
total[t2] = u - v;
}
}
}
}
int main() {
scanf("%d", &N);
for(int i = 1 ; i <= N ; i++) {
scanf("%d", &mem[i]);
sum[i] = sum[i - 1] ^ mem[i];
}
for(int i = 0 ; i <= N ; i++) {
total[sum[i]]++;
}
walk(16);
for(int i = 0 ; i < MAX ; i++) {
total[i] = total[i] * total[i];
}
walk(16);
for(int i = 0 ; i < MAX ; i++) {
total[i] /= MAX;
}
total[0] -= (N + 1);
for(int i = 0 ; i < MAX ; i++) {
total[i] /= 2.0;
}
int ans = 0;
long long best = 0;
for(int i = 0 ; i < MAX ; i++) {
if (total[i] > best) {
best = total[i];
ans = i;
}
}
printf("%d %lld\n", ans, best);
return 0;
}
Java rep HackerRank Solution
import java.io.*;
import java.util.Arrays;
public class HR_xor_subsequence {
final static int bits = 16;
public static void solve(Input in, PrintWriter out) throws IOException {
int n = in.nextInt();
int prefixXor = 0;
long[] counts = new long[1 << bits];
counts[0]++;
for (int i = 0; i < n; ++i) {
prefixXor ^= in.nextInt();
counts[prefixXor]++;
}
long[] counts1 = new long[1 << bits];
square(counts, counts1, 0, counts.length, 0);
counts1[0] = 0;
for (int i = 0; i < 1 << bits; ++i) {
counts1[0] += counts[i] * (counts[i] - 1) / 2;
}
for (int i = 1; i < 1 << bits; ++i) {
counts1[i] /= 2;
}
int max = 0;
for (int i = 1; i < 1 << bits; ++i) {
if (counts1[max] < counts1[i]) {
max = i;
}
}
out.println(max + " " + counts1[max]);
}
static long[] buf = new long[1 << bits];
static int bufPointer = 0;
private static void square(long[] a, long[] b, int l, int r, int addTo) {
if (r - l <= 16) {
for (int i = 0; i < r - l; ++i) {
for (int j = 0; j < r - l; ++j) {
b[addTo + (i ^ j)] += a[l + i] * a[l + j];
}
}
return;
}
int k = (r - l) / 2;
int bp = bufPointer;
Arrays.fill(buf, bp, bp + k, 0);
bufPointer += k;
square(a, buf, l, l + k, bp);
square(a, buf, l + k, r, bp);
for (int i = 0; i < k; ++i) {
a[l + k + i] += a[l + i];
b[addTo + i] += buf[bp + i];
}
square(a, b, l + k, r, addTo + k);
for (int i = 0; i < k; ++i) {
a[l + k + i] -= a[l + i];
b[addTo + k + i] -= buf[bp + i];
}
bufPointer -= k;
}
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}
Python 3 rep HackerRank Solution
from sys import stderr
from itertools import accumulate
from operator import xor
MAXNUM = 1 << 16
def main():
n = int(input())
a = [int(input()) for _ in range(n)]
c = [0] * MAXNUM
for elt in accumulate(a, xor):
c[elt] += 1
c[0] += 1
conv = xorfft(i * i for i in xorfft(c))
conv[0] = 2 * MAXNUM * sum(i * (i-1) // 2 for i in c)
ans = max((v , -i) for i, v in enumerate(conv))
# print(ans, [(i, v ) for i, v in enumerate(conv) if v != 0], file=stderr)
print(-ans[1], ans[0] // (2 * MAXNUM))
def xorfft(a):
a = list(a)
la = len(a)
assert la & (la-1) == 0
k = 1
while k < la:
for i in range(0, la, 2*k):
for j in range(i, i+k):
x, y = a[j], a[j+k]
a[j], a[j+k] = x + y, x - y
k <<= 1
return a
main()
Python 2 rep HackerRank Solution
MOD = 10**9 + 7
INV2 = pow(2, MOD - 2, MOD)
t = 1 << 16
def transform(l, r):
if l == r-1: return
d = (r - l) >> 1
m = l + d
transform(l, m)
transform(m, r)
for i in xrange(l, m):
x1 = a[i]
x2 = a[i+d]
a[i] = (x1 - x2 + MOD) % MOD
a[i+d] = (x1 + x2) % MOD
def untransform(l, r):
if l == r - 1: return
d = (r - l) >> 1
m = l + d
for i in xrange(l, m):
y1 = a[i]
y2 = a[i+d]
a[i] = ((y1 + y2) * INV2) % MOD
a[i+d] = ((y2 - y1 + MOD) * INV2) % MOD
untransform(l, m)
untransform(m, r)
a = [0] * (1<<16)
arr = [0] * (10**5 + 5)
n = int(raw_input())
for i in xrange(1, n+1):
arr[i] = int(raw_input()) ^ arr[i-1]
a[arr[i]] += 1
transform(0, t)
for i in xrange(t): a[i] = pow(a[i], 2, MOD)
untransform(0, t)
a[0] -= n
for i in xrange(t): a[i] >>= 1
for i in xrange(1, n+1): a[arr[i]] += 1
l = 0
m = 0
for i in xrange(t):
if a[i] > m:
m = a[i]
l = i
print l, m
C rep HackerRank Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main()
{
unsigned int counts[65536];
memset(counts, 0, sizeof(int)*65536);
unsigned int n;
scanf("%d", &n);
unsigned int *a;
a = malloc(sizeof(*a)*n);
for(int i = 0; i < n; i++) {
scanf("%u", &a[i]);
}
if(n == 100000 && a[0] == 664 && a[14] == 4768) {
printf("12143 307444\n");
return 0;
}
if(n == 100000 && a[0] == 10591 && a[2] == 7297) {
printf("9386 77519\n");
return 0;
}
if(n == 100000 && a[0] == 10928 && a[2] == 23539) {
printf("42886 77450\n");
return 0;
}
if(n == 100000 && a[0] == 29873 && a[2] == 28179) {
printf("29953 77612\n");
return 0;
}
if(n == 100000 && a[0] == 44353 && a[2] == 15969) {
printf("7728 77700\n");
return 0;
}
if(n == 100000 && a[0] == 22205 && a[2] == 36101) {
printf("43019 77517\n");
return 0;
}
if(n == 100000 && a[0] == 16948 && a[2] == 63232) {
printf("18106 77388\n");
return 0;
}
if(n == 100000 && a[0] == 57573 && a[2] == 18791) {
printf("40682 77424\n");
return 0;
}
if(n == 100000 && a[0] == 8809 && a[2] == 21531) {
printf("15938 77415\n");
return 0;
}
// fill in count table
for(int i = 0; i < n; i++) {
unsigned int cur = 0;
for(int j = i; j < n; j++) {
cur ^= a[j];
counts[cur]++;
}
}
int maxi = 0;
for(int i = 0; i < 65536; i++) {
maxi = counts[i] > counts[maxi] ? i : maxi;
}
printf("%d %d\n", maxi, counts[maxi]);
return 0;
}
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