Dadas N – 1 diferencias entre dos elementos consecutivos de una array que contiene N números que están en el rango de 1 a N. La tarea es determinar la array original usando las diferencias dadas. Si es posible, imprima la array, de lo contrario, imprima -1 .
Ejemplos:
Entrada: diff[] = {2, -3, 2}
Salida: arr[] = {2, 4, 1, 3}
4 – 2 = 2
1 – 4 = -3
3 – 1 = 2
Entrada: diff[] = {2, 2, 2}
Salida: -1
Enfoque: Ya que queremos generar permutaciones en el rango [1, n] y cada dígito debe ocurrir solo una vez. Tomar como ejemplo,
arr[] = {2, -3, 2}
Aquí, P2 – P1 = 2 , P3 – P2 = -3 , P4 – P3 = 2 .
Sea P1 = x entonces P2 = x + 2 , P3 = P2 – 3 = x + 2 – 3 = x – 1 , P4 = P3 + 2 = x – 1 + 2 = x + 1.
Entonces, P1 = x , P2 = X + 2 , P3 = X – 1 , P4 = X + 1 .
Ahora, dado que queremos una permutación de 1 a n , las P[i] que obtenemos arriba también deben satisfacer la condición. Dado que el valor debe satisfacerse para todos y cada uno de los x , por lo tanto, para simplificar, tomamos x = 0.
Ahora, P1 = 0, P2 = 2, P3 = -1, P4 = 1.
Ordenaremos los p[i] , después de ordenar la diferencia consecutiva entre cada elemento debe ser igual a 1. Si en cualquier índice encontramos un elemento p[i] tal que p[i] – p[i – 1] != 1 entonces no es posible generar la permutación. Para generar la permutación final, realizaremos un seguimiento de los índices, podemos usar map o unordered_map para hacerlo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the required permutation
void findPerm(int n, vector<int>& differences)
{
vector<int> ans;
ans.clear();
ans.push_back(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for (int i = 0; i <= n - 2; ++i) {
int diff = differences[i];
x = x + diff;
ans.push_back(x);
}
// Preserve the original array
vector<int> anss = ans;
sort(ans.begin(), ans.end());
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for (int i = 1; i <= n - 1; ++i) {
int res = ans[i] - ans[i - 1];
if (res != 1) {
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0) {
cout << -1;
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else {
unordered_map<int, int> mpp;
mpp.clear();
int j = 1;
vector<int> value_at_index;
for (auto& x : ans) {
mpp[x] = j;
++j;
}
for (auto& x : anss) {
value_at_index.push_back(mpp[x]);
}
for (auto& x : value_at_index) {
cout << x << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
vector<int> differences;
differences.push_back(2);
differences.push_back(-3);
differences.push_back(2);
int n = differences.size() + 1;
findPerm(n, differences);
return 0;
}
Java
// Java program to implement the above approach
import java.util.*;
class GFG
{
// Function to print the required permutation
static void findPerm(int n, Vector<Integer> differences)
{
Vector<Integer> ans = new Vector<Integer>();
ans.clear();
ans.add(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for (int i = 0; i <= n - 2; ++i)
{
int diff = differences.get(i);
x = x + diff;
ans.add(x);
}
// Preserve the original array
Vector<Integer> anss = new Vector<Integer>();
for(Integer obj:ans)
anss.add(obj);
Collections.sort(ans);
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for (int i = 1; i <= n - 1; ++i)
{
int res = ans.get(i) - ans.get(i - 1);
if (res != 1)
{
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0)
{
System.out.print(-1);
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else
{
Map<Integer,Integer> mpp = new HashMap<>();
mpp.clear();
int j = 1;
Vector<Integer> value_at_index = new Vector<Integer>();
for (Integer x1 : ans)
{
mpp.put(x1, j);
++j;
}
for (Integer x2 : anss)
{
value_at_index.add(mpp.get(x2));
}
for (Integer x3 : value_at_index)
{
System.out.print(x3 + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
Vector<Integer> differences = new Vector<Integer>();
differences.add(2);
differences.add(-3);
differences.add(2);
int n = differences.size() + 1;
findPerm(n, differences);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
# Function to print the required permutation
def findPerm(n, differences):
ans = []
ans.append(0)
# Take x = 0 for simplicity
x = 0
# Calculate athe differences
# and store it in a vector
for i in range(n - 1):
diff = differences[i]
x = x + diff
ans.append(x)
# Preserve the original array
anss = ans
ans = sorted(ans)
flag = -1
# Check if athe consecutive elements
# have difference = 1
for i in range(1, n):
res = ans[i] - ans[i - 1]
if (res != 1):
flag = 0
# If consecutive elements don't have
# difference 1 at any position then
# the answer is impossible
if (flag == 0):
print("-1")
return
# Else store the indices and values
# at those indices in a map
# and finainty print them
else:
mpp = dict()
j = 1
value_at_index = []
for x in ans:
mpp[x] = j
j += 1
for x in anss:
value_at_index.append(mpp[x])
for x in value_at_index:
print(x, end = " ")
print()
# Driver code
differences=[]
differences.append(2)
differences.append(-3)
differences.append(2)
n = len(differences) + 1
findPerm(n, differences)
# This code is contributed by mohit kumar
C#
// C# program to implement the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the required permutation
static void findPerm(int n, List<int> differences)
{
List<int> ans = new List<int>();
ans.Clear();
ans.Add(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for(int i = 0; i <= n - 2; ++i)
{
int diff = differences[i];
x = x + diff;
ans.Add(x);
}
// Preserve the original array
List<int> anss = new List<int>();
foreach(int obj in ans)
anss.Add(obj);
ans.Sort();
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for(int i = 1; i <= n - 1; ++i)
{
int res = ans[i] - ans[i - 1];
if (res != 1)
{
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0)
{
Console.Write(-1);
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else
{
Dictionary<int,
int> mpp = new Dictionary<int,
int>();
mpp.Clear();
int j = 1;
List<int> value_at_index = new List<int>();
foreach (int x1 in ans)
{
mpp.Add(x1, j);
++j;
}
foreach (int x2 in anss)
{
value_at_index.Add(mpp[x2]);
}
foreach (int x3 in value_at_index)
{
Console.Write(x3 + " ");
}
Console.WriteLine();
}
}
// Driver code
public static void Main(String[] args)
{
List<int> differences = new List<int>();
differences.Add(2);
differences.Add(-3);
differences.Add(2);
int n = differences.Count + 1;
findPerm(n, differences);
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// Javascript program to implement the above approach
// Function to print the required permutation
function findPerm(n,differences)
{
let ans = [];
ans.push(0);
// Take x = 0 for simplicity
let x = 0;
// Calculate aint the differences
// and store it in a vector
for (let i = 0; i <= n - 2; ++i)
{
let diff = differences[i];
x = x + diff;
ans.push(x);
}
// Preserve the original array
let anss = [];
for(let obj = 0; obj < ans.length; obj++)
anss.push(ans[obj]);
ans.sort(function(a,b){return a-b;});
let flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for (let i = 1; i <= n - 1; ++i)
{
let res = ans[i] - ans[i - 1];
if (res != 1)
{
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0)
{
document.write(-1);
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else
{
let mpp = new Map();
let j = 1;
let value_at_index = [];
for (let x1 = 0; x1 < ans.length; x1++)
{
mpp.set(ans[x1], j);
++j;
}
for (let x2 = 0; x2 < anss.length; x2++)
{
value_at_index.push(mpp.get(anss[x2]));
}
for (let x3 = 0; x3 < value_at_index.length; x3++)
{
document.write(value_at_index[x3] + " ");
}
document.write("<br>");
}
}
// Driver code
let differences =[];
differences.push(2);
differences.push(-3);
differences.push(2);
let n = differences.length + 1;
findPerm(n, differences);
// This code is contributed by unknown2108.
</script>
Producción:
2 4 1 3
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(n)