## Facebook Interview Question

Country: United States

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SOLUTION:

public void nextPermutation(int[] nums) {
if(nums.length <= 1)  return;

int i = nums.length - 1;
while(i >= 1 && nums[i] <= nums[i-1]) i--; //find first number which is smaller than it's after number
if(i!=0) swap(nums,i-1); //if the number exist,which means that the nums not like{5,4,3,2,1}
reverse(nums,i);
}

private void swap(int[] a,int i){
for(int j = a.length - 1;j > i;j--){
if(a[j] > a[i]){
int t = a[j];
a[j] = a[i];
a[i] = t;
break;
}
}
}

private void reverse(int[] a,int i){//reverse the number after the number we have found
int first = i;
int last = a.length - 1;
while(first<last){
int t = a[first];
a[first] = a[last];
a[last] = t;
first++;
last--;
}
}

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Looks like your method will take int array. But in the above problem it says int number. How does it work ?

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def next_permutation(x):
v=list(str(x))
n=len(v)
vr=[]
if n>1:
for i in range(1,n):
pc=v[-1*i]
pn=v[-1*(i+1)]
if pc>pn:
vr.append(pc)
vr.sort()
for j in vr:
if j>pn:
vr.append(str(pn))
vr.sort()
v[-1*(i+1)]=str(j)
vr.remove(str(j))
for counter in range(0, len(vr)):
z=vr[0]
v[-1*(i+1)+1+counter]=z
vr.remove(z)
x=''.join(v)
x=int(x)
return x
else:
vr.append(str(pc))
return x

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def smallest_greater_perm(x):
smallest_perm=0
list_digits=[]
while x>0:
list_digits.append(x % 10)
x//=10
if x==0: break
if list_digits==sorted(list_digits): return print('No greater permutation exists')
else:
for i,item in enumerate(list_digits):
if item < list_digits[0]:
list_digits[0],list_digits[i]=list_digits[i],list_digits[0]
break
for i,item in enumerate(list_digits):
smallest_perm+=item*10**i
return smallest_perm

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private static void findNextPermutation(char[] numArr) {
if(numArr.length == 0) {
return;
}
// Find the max i, so that K(i-1) < ki
// Find the max j, so that K(i-1) < kj
// Swap Ki and K(i-1)
// reverse from Ki, ki+1.... kn;
int i = numArr.length -2;
while(i >=0 && numArr[i+1] <= numArr[i]) {
i--;
}
if(i >=0) {
int j = numArr.length -1;
while(j >=0 && numArr[j] <= numArr[i]) {
j--;
}
swap(numArr, i, j);
}
reverse(numArr, i+1);

for(char c : numArr) {
System.out.print(c + ", ");
}
}

private static void reverse(char[] numArr, int start) {

int i = start; int j = numArr.length -1;
while(i <j) {
swap(numArr,i,j);
i++;
j--;
}
}

private static void swap(char[] numArr, int i, int j) {
char temp = numArr[i];
numArr[i] = numArr[j];
numArr[j] = temp;
}

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public class Permutation {
public static void main(String[] args) {
int num = 231145;
permutation(num);
}

public static void permutation(int num) {
List<String> tokenize = new ArrayList<String>();
while (num > 0) {
num /= 10;
}

String[] sortedArr = tokenize.toArray(new String[0]);
Arrays.sort(sortedArr);
System.out.println();
for (int i = 1; i < sortedArr.length; i++) {
if (!sortedArr[i].equals(sortedArr[i - 1])) {
String swap = sortedArr[i];
sortedArr[i] = sortedArr[i - 1];
sortedArr[i - 1] = swap;
break;
}
}

for (int i = sortedArr.length-1; i >= 0; i--) {
System.out.print(sortedArr[i]);
}

}

}

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public void nextPermutation(int[] a) {
int n = a.length;

int i = n - 1;
while(i-1 >= 0 && a[i-1] >= a[i]) i--;  // Find the smallest i where  a[i..n-1] is reverse sorted

if(i != 0) {                           // if i == 0, the input is largest permutation, so just need to sort array
int j = n - 1;
while(a[i-1] >= a[j]) j--;         // find the smallest j where a[i-1] < a[j]

swap(a, i-1, j);                  // swap a[i-1] and a[j]
}

Arrays.sort(a, i, n);                  // sort a[i-1..n-1]
}

void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}

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public void nextPermutation(int[] a) {
int n = a.length;

int i = n - 1;
while(i-1 >= 0 && a[i-1] >= a[i]) i--;  // Find the smallest i where  a[i..n-1] is reverse sorted

if(i != 0) {                           // if i == 0, the input is largest permutation, so just need to sort array
int j = n - 1;
while(a[i-1] >= a[j]) j--;         // find the smallest j where a[i-1] < a[j]

swap(a, i-1, j);                  // swap a[i-1] and a[j]
}

Arrays.sort(a, i, n);                  // sort a[i-1..n-1]
}

void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}

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O(n log n) solution in C++ using DP.

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

vector<int> digits;

const int INF = 1e8;

vector<vector<pair<int, int>>> value;
pair<int, int> larger(int d, int i) {
auto & res = value[d][i];
if(res.first != -1) {
return res;
}
if(i >= digits.size()) {
return res = {INF, -1};
}
auto p = larger(d, i+1);
if(digits[i+1] > d && digits[i+1] < p.first) {
return res = {digits[i+1], i+1};
}
return res = p;
}

void debug() {
for( int d : digits ) {
cout << d << " ";
}
cout << '\n';
}

bool next_permutation(int i) {
if(i >= digits.size()) {
return false;
}
value.assign(10, vector<pair<int,int>>(digits.size()+2, {-1, -1}));
auto p = larger(digits[i], i);
if(p.second != -1) {

swap(digits[i], digits[p.second]);

sort(digits.begin()+i+1, digits.end());
return true;
}
return false;
}

bool next_permutation() {
for(int i = digits.size() -1; i >= 0; --i) {
if(next_permutation(i)) return true;
}
return false;
}

int main() {

digits = {3, 4, 1, 5, 2};

while(next_permutation()) {
for( int d : digits ) {
cout << d << " ";
}
cout << '\n';
}
return 0;
}

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from bisect import bisect_right, insort

def next_smallest_permutation(n):
ns = str(n)
l = len(ns)
arr = [ns[-1]]
for i in range(l-2, -1, -1):
insort(arr, ns[i])
if ns[i] < ns[i+1]:
j = bisect_right(arr, ns[i])
anss = ns[:i] + arr[j] + ''.join(arr[:j]) + ''.join(arr[j+1:])
return int(anss)
#Retuns the rolled over permutation i.e smallest parmutation if it is already the largest
return ''.join(arr)

# Tests
print(next_smallest_permutation(825))
print(next_smallest_permutation(82552))
print(next_smallest_permutation(95432))

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public static int nextPer(int num) {

int[] arr= numToArr(num);
int lastSwap=0;
for(int i=arr.length-1;i>lastSwap;i--){
for(int j=i;j>lastSwap;j--){
if(arr[i]>arr[j]){
swap(arr,j,i);
lastSwap=j;
}
}
}

return arrToNum(arr);

}

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package practice2018;

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class NextPermutation {
public static void main(String[] args){
String[][] testcases = {
{"12", "21"},
{"315","351"},
{"583","835"},
{"12389","12398"},
{"34722641","34724126"}
};
for(int i = 0; i < testcases.length; i++){
try {
String actual = next(testcases[i][0]);
System.out.println("test " + testcases[i][0] + " actual " + testcases[i][1] + " " + actual);
} catch (Exception e){
System.out.println("test " + testcases[i][0] + " " + e.getMessage());
}
}
}

public static String next(String inp) throws Exception {
char[] digits = inp.toCharArray();
int n = digits.length;

for (int i = n-1; i> 0; i--){
if(digits[i-1] < digits[i]){
// find immediate next permutation for substring(digits,i-1)
List<Character> a = new ArrayList<Character>();
for (int j = i-1; j < n; j++){
}
Collections.sort(a);

// find smallest digit larger than digits[i-1] in digits[i, ..]
String partial = "";
for (int j = 0; j < a.size(); j++){
if (a.get(j) > digits[i-1]){
partial += a.get(j);

for (int k = 0; k < a.size(); k++){
if (k != j){
partial += a.get(k);
}
}
break;
}
}

String result = "";
for (int j = 0; j < i-1; j++){
result += digits[j];
}
result += partial;
return result;
}
}
throw new Exception("At maximum, cannt find next");
}
}

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of 0 vote

int next_permutation(int v) {
std::string s = std::to_string(v);
int i = s.length() - 1;
while (i > 0) {
if (s[i - 1] > s[i]) {
--i;
}
else {
int min = i;
for (int j = i + 1; j < s.length(); ++j) {
if (s[j] > s[i - 1] && s[j] < s[min]) {
min = j;
}
}
std::swap(s[min], s[i - 1]);
std::sort(s.begin() + i, s.end());
break;
}
}
return atoi(s.c_str());
}

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of 1 vote

int greater(int num)
{
string str = to_string(num);
int n = str.size();
int i = 0, j=0;

for(i=n-1; i>-1; i--)
{
for (j=i-1; j>-1; j--)
{
if (str[i] > str[j])
{
str = str.substr(0, j) + str[i] + str.substr(j, i - j) + str.substr(i+1, n-(i+1));
sort(str.begin() + j + 1, str.end());
return stoi(str);
}
}
}
return -1;
}

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