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Remove One Element CodeChef Solution
Problem -Remove One Element CodeChef Solution
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Remove One Element CodeChef Solution in C++17
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
#define f(i,x,n) for(ll i=x;i<n;i++)
#define all(a) a.begin() , a.end()
#define fast_io() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define print(a) cout << a << ' '
#define println(a) cout << a << '\n'
#define line() cout << '\n'
inline void solve() {
int n; cin >> n;
vi a(n),b(n-1);
f(i,0,n) cin >> a[i];
f(i,0,n-1) cin >> b[i];
sort(all(a));
sort(all(b));
map<int,int> freq;
int d1,d2;
f(i,0,n-1) {
d1 = b[i] - a[i];
d2 = b[i] - a[i+1];
if( d1 > 0 ) freq[d1]++;
if( d2 > 0 ) freq[d2]++;
}
for(auto f:freq ) {
if(f.second == n-1) {
println(f.first);
return;
}
}
}
signed main() {
fast_io();
int t = 1;
cin >> t;
while(t--) solve();
return 0;
}Remove One Element CodeChef Solution in C++14
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define trav(a, x) for(auto &a:x)
#define rep(i, b) for(int i=0; i<b; i++)
#define in(n) int n; cin>>n;
#define vi vector<int>
#define vc vector<char>
#define vs vector<string>
#define vvi vector<vi>
#define vpii vector<pair<int,int>>
#define mii map<int,int>
#define mci map<char,int>
#define msi map<string,int>
#define pb push_back
#define pob pop_back
#define fi first
#define se second
#define YES cout << "YES" << endl;
#define yes cout << "yes" << endl;
#define Yes cout << "Yes" << endl;
#define NO cout << "NO" << endl;
#define no cout << "no" << endl;
#define No cout << "No" << endl;
#define minus1 cout<<-1<<endl;
#define zero cout<<0<<endl;
#define coutt(x) cout<<x<<endl;
#define cott(x) cout<<x<<" ";
#define end cout << endl;
#define mod 998244353
void solve()
{
int n;
cin>>n;
int arr[n];
int brr[n-1];
rep(i,n)cin>>arr[i];
rep(i,n-1)cin>>brr[i];
sort(arr,arr+n);
sort(brr,brr+n-1);
mii m;
for(int i=0;i<n-1;i++)
{
m[brr[i]-arr[i]]++;
m[brr[i]-arr[i+1]]++;
}
int value = -1;
int val = 0;
for(auto i:m)
{
if(i.se>val && i.fi>0)
{
value = i.fi;
val = i.se;
}
}
cout<<value<<endl;
}
int32_t main()
{
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}Remove One Element CodeChef Solution in PYTH 3
'''
bob's array is 1 smaller
Step 1: Sort the 2 arrays
Step 2: Either x = bobArray[0] - aliceArray[0]
or
x = bobArray[0] - aliceArray[1]
Both cannot be false
Since we want smallest value of x, try checking for 2nd case
If not, 1st case should be true.
'''
def decodeX(aliceArray, bobArray):
if not isinstance(aliceArray, list) or not isinstance(bobArray, list):
raise ValueError
aliceArray.sort()
bobArray.sort()
deltaX1 = bobArray[0] - aliceArray[0]
deltaX2 = bobArray[0] - aliceArray[1]
if deltaX2 <= 0:
return deltaX1
for i in range(0, len(bobArray)):
deltaX = bobArray[i] - aliceArray[i+1]
if deltaX != deltaX2:
return deltaX1
return deltaX2
def solve():
# ignore N
input()
aliceArray = list(map(int, input().split()))
bobArray = list(map(int, input().split()))
x = decodeX(aliceArray, bobArray)
print(x)
def main():
test_cases = int(input())
for t in range(test_cases):
solve()
if __name__ == "__main__":
main()Remove One Element CodeChef Solution in C
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define lli long long int
#define li long int
#define inf 10000
int max(int a, int b)
{
int m = a;
if(b>a)
m = b;
return m;
}
int min(int a, int b)
{
int m = a;
if(b<a)
m = b;
return m;
}
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(lli arr[], lli l, lli m, lli r)
{
lli i, j, k;
lli n1 = m - l + 1;
lli n2 = r - m;
/* create temp arrays */
lli L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(lli arr[], lli l, lli r)
{
if (l < r) {
// Same as (l+r)/2, but avoids overflow for
// large l and h
lli m = l + (r - l) / 2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
void program()
{
/*
sort them
just take diff of first n-1 (bi - ai)
if n >= 3 : two diff must be equal, take them
if n = 2 : take second diff
*/
lli n;
scanf("%lld", &n);
lli i;
lli A[n], B[n-1];
for ( i = 0; i < n; i++)
{
scanf("%lld", &A[i]);
}
for ( i = 0; i < n-1; i++)
{
scanf("%lld", &B[i]);
}
mergeSort(A, 0, n-1);
mergeSort(B, 0, n-2);
if(n == 2)
{
if(A[1] >= B[0])
printf("%lld\n", B[0]-A[0]);
else
printf("%lld\n", B[0]-A[1]);
return;
}
lli d1 = B[n-2] - A[n-1];
lli d2 = B[n-3] - A[n-2];
if(d1 == d2 && d1>0)
{
printf("%lld\n", d1);
return;
}
//eliminate A[n-2]
lli d2_ = B[n-3] - A[n-3];
if(d1 == d2_ && d1>0)
{
printf("%lld\n", d1);
return;
}
//eliminate A[n-1]
lli d1_ = B[n-2] - A[n-2];
if(d1_ == d2_)
{
printf("%lld\n", d1_);
return;
}
}
int main(void)
{
long long int T;
scanf("%lld", &T);
//printf("%lld\n", T);
// T = 1;
for (long long int i = 0; i < T; i++)
{
program();
}
return 0;
}
Remove One Element CodeChef Solution in JAVA
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int disha=sc.nextInt();
while(disha-->0){
int cnt=0;
int nocnt=0;
int n=sc.nextInt();
int arr[]=new int[n];
int brr[]=new int[n-1];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
for(int i=0;i<n-1;i++){
brr[i]=sc.nextInt();
}
Arrays.sort(arr);
Arrays.sort(brr);
int diff1=brr[0]-arr[0];
int diff2=brr[0]-arr[1];
int cnt1=0;
int cnt2=0;
for(int i=0;i<brr.length;i++){
if(brr[i]-arr[i]>0){
if(brr[i]-arr[i]==diff1)cnt1++;
if(brr[i]-arr[i]==diff2)cnt2++;
}
if(brr[i]-arr[i+1]>0){
if(brr[i]-arr[i+1]==diff1)cnt1++;
if(brr[i]-arr[i+1]==diff2)cnt2++;
}
}
if(diff1<=0)
System.out.println(diff2);
else if(diff2<=0)
System.out.println(diff1);
else if(cnt1 > cnt2 && diff2 > 0) {
System.out.println(diff1);
}
else {
System.out.println(diff2);
}
}
}
}Remove One Element CodeChef Solution in PYPY 3
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a.sort()
b.sort()
m = min(b)-min(a)
M = max(b)-max(a)
# the answer is either m or M
if M <= 0:
print(m)
elif n == 2:
print(min(m, M))
elif b[0]-a[1] == M:
print(M)
else:
print(m)Remove One Element CodeChef Solution in PYTH
# cook your code here
import collections
T = input()
while T>0:
N = input()
A = sorted(map(int,raw_input().split()))
B = sorted(map(int,raw_input().split()))
if N == 2:
if A[0]<B[0]: print(min(B[0] - A[0],B[0]-A[1]) if B[0]>A[1] else B[0]-A[0])
else: print(B[0]-A[1])
else:
if B[0]-A[1] == B[1]-A[2] and B[0]-A[1]!=0: print(B[0]-A[1])
else:print(B[0]-A[0])
T-=1Remove One Element CodeChef Solution in C#
using System;
using System.Linq;
using System.Collections.Generic;
using System.Text.RegularExpressions;
using System.Text;
using System.Numerics;
public class Test
{
public static void Main()
{
var WS = new StringBuilder();
int A = GetInt();
int N;
while(A-- != 0)
{
var n = GetInt();
var a = GetIntArray().ToList();
var b = GetIntArray().ToList();
var originalMin = a.Min();
var originalSecondMin = a.Where(x => x > originalMin).Min();
var newMin = b.Min();
var x1 = newMin - originalMin;
var x2 = newMin - originalSecondMin;
var originalSet = new HashSet<int>(a);
var isValid2 = x2 > 0 && b.All(x => originalSet.Contains(x - x2));
if(isValid2) Console.WriteLine(x2);
else Console.WriteLine(x1);
}
Console.WriteLine(WS.ToString());
// your code goes here
}
public static void Writex(string s) => (Console.WriteLine(s));
public static int GetInt() => int.Parse(Console.ReadLine());
public static long GetLong() => long.Parse(Console.ReadLine());
public static string GetString() => (Console.ReadLine());
public static string[] GetStringArray() => Console.ReadLine().Trim().Split(' ');
public static int[] GetIntArray() => Console.ReadLine().Trim().Split(' ').Select(int.Parse).ToArray();
public static long[] GetLongArray() => Console.ReadLine().Trim().Split(' ').Select(long.Parse).ToArray();
}Remove One Element CodeChef Solution in GO
package main
import (
"bufio"
"bytes"
"fmt"
"os"
"sort"
)
func main() {
reader := bufio.NewReader(os.Stdin)
tc := readNum(reader)
var buf bytes.Buffer
for tc > 0 {
tc--
n := readNum(reader)
A := readNNums(reader, n)
B := readNNums(reader, n-1)
res := solve(n, A, B)
buf.WriteString(fmt.Sprintf("%d\n", res))
}
fmt.Print(buf.String())
}
func readInt(bytes []byte, from int, val *int) int {
i := from
sign := 1
if bytes[i] == '-' {
sign = -1
i++
}
tmp := 0
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + int(bytes[i]-'0')
i++
}
*val = tmp * sign
return i
}
func readNum(reader *bufio.Reader) (a int) {
bs, _ := reader.ReadBytes('\n')
readInt(bs, 0, &a)
return
}
func readTwoNums(reader *bufio.Reader) (a int, b int) {
res := readNNums(reader, 2)
a, b = res[0], res[1]
return
}
func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
res := readNNums(reader, 3)
a, b, c = res[0], res[1], res[2]
return
}
func readNNums(reader *bufio.Reader, n int) []int {
res := make([]int, n)
x := 0
bs, _ := reader.ReadBytes('\n')
for i := 0; i < n; i++ {
for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
x++
}
x = readInt(bs, x, &res[i])
}
return res
}
func readUint64(bytes []byte, from int, val *uint64) int {
i := from
var tmp uint64
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + uint64(bytes[i]-'0')
i++
}
*val = tmp
return i
}
func solve(n int, A []int, B []int) int {
sort.Ints(A)
sort.Ints(B)
x := B[0] - A[0]
if x > 0 {
i := 1
j := 1
var cnt int
for i < n-1 {
if j < n && B[i]-A[j] == x {
i++
j++
continue
}
j++
cnt++
if j < n && B[i]-A[j] == x {
i++
j++
continue
}
break
}
if j < n {
cnt++
}
if i < n-1 || cnt != 1 {
// bad choice
x = -1
}
}
y := B[0] - A[1]
if y > 0 {
i := 1
j := 2
for i < n-1 && j < n && B[i]-A[j] == y {
i++
j++
}
if i < n-1 {
y = -1
}
}
if x > 0 && y > 0 {
return min(x, y)
}
if x > 0 {
return x
}
return y
}
func min(a, b int) int {
if a < b {
return a
}
return b
}Remove One Element CodeChef Solution Review:
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