Restore a permutation from the given helper array
Last Updated :
01 Mar, 2022
Given an array Q[] of size N - 1 such that each Q[i] = P[i + 1] - P[i] where P[] is the permutation of the first N natural numbers, the task is to find this permutation. If no valid permutation P[] can be found then print -1.
Examples:
Input: Q[] = {-2, 1}
Output: 3 1 2
Input: Q[] = {1, 1, 1, 1}
Output: 1 2 3 4 5
Approach: This is a mathematical algorithmic question. Lets denote P[i] = x. Therefore P[i + 1] = P[i] + (P[i + 1] - P[i]) = x + Q[i] (Since Q[i] = P[i + 1] - P[i]).
Therefore, P[i + 2]= P[i] + (P[i + 1] - P[i]) + (P[i + 2] - P[i + 1]) = x + Q[i] + Q[i + 1]. Observe, the pattern forming here. P is nothing but [x, x + Q[1], x + Q[1] + Q[2] + ... + x + Q[1] + Q[2] + ... + Q[n - 1]] where x = P[i] which is still unknown.
Lets have a permutation P' where P'[i] = P[i] - x. Therefore, P' = [0, Q[1], Q[1] + Q[2], Q[1] + Q[2] + Q[3], ..., Q[1] + Q[2] + ... + Q[n - 1]].
To find x, lets find the smallest element in P'. Let it be P'[k]. Therefore, x = 1 - P'[k]. This is because, the original permutation P has integers from 1 to n and so 1 can be the minimum element in P. After finding x, add x to each P' to get the original permutation P.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the required permutation
void findPerm(int Q[], int n)
{
int minval = 0, qsum = 0;
for (int i = 0; i < n - 1; i++) {
// Each element in P' is like a
// cumulative sum in Q
qsum += Q[i];
// minval is the minimum
// value in P'
if (qsum < minval)
minval = qsum;
}
vector<int> P(n);
P[0] = 1 - minval;
// To check if each entry in P
// is from the range [1, n]
bool permFound = true;
for (int i = 0; i < n - 1; i++) {
P[i + 1] = P[i] + Q[i];
// Invalid permutation
if (P[i + 1] > n || P[i + 1] < 1) {
permFound = false;
break;
}
}
// If a valid permutation exists
if (permFound) {
// Print the permutation
for (int i = 0; i < n; i++) {
cout << P[i] << " ";
}
}
else {
// No valid permutation
cout << -1;
}
}
// Driver code
int main()
{
int Q[] = { -2, 1 };
int n = 1 + (sizeof(Q) / sizeof(int));
findPerm(Q, n);
return 0;
}
// Java implementation of the approach
class GFG
{
// Function to find the required permutation
static void findPerm(int Q[], int n)
{
int minval = 0, qsum = 0;
for (int i = 0; i < n - 1; i++)
{
// Each element in P' is like a
// cumulative sum in Q
qsum += Q[i];
// minval is the minimum
// value in P'
if (qsum < minval)
minval = qsum;
}
int []P = new int[n];
P[0] = 1 - minval;
// To check if each entry in P
// is from the range [1, n]
boolean permFound = true;
for (int i = 0; i < n - 1; i++)
{
P[i + 1] = P[i] + Q[i];
// Invalid permutation
if (P[i + 1] > n || P[i + 1] < 1)
{
permFound = false;
break;
}
}
// If a valid permutation exists
if (permFound)
{
// Print the permutation
for (int i = 0; i < n; i++)
{
System.out.print(P[i]+ " ");
}
}
else
{
// No valid permutation
System.out.print(-1);
}
}
// Driver code
public static void main(String[] args)
{
int Q[] = { -2, 1 };
int n = 1 + Q.length;
findPerm(Q, n);
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation of the approach
# Function to find the required permutation
def findPerm(Q, n) :
minval = 0; qsum = 0;
for i in range(n - 1) :
# Each element in P' is like a
# cumulative sum in Q
qsum += Q[i];
# minval is the minimum
# value in P'
if (qsum < minval) :
minval = qsum;
P = [0]*n;
P[0] = 1 - minval;
# To check if each entry in P
# is from the range [1, n]
permFound = True;
for i in range(n - 1) :
P[i + 1] = P[i] + Q[i];
# Invalid permutation
if (P[i + 1] > n or P[i + 1] < 1) :
permFound = False;
break;
# If a valid permutation exists
if (permFound) :
# Print the permutation
for i in range(n) :
print(P[i],end=" ");
else :
# No valid permutation
print(-1);
# Driver code
if __name__ == "__main__" :
Q = [ -2, 1 ];
n = 1 + len(Q) ;
findPerm(Q, n);
# This code is contributed by AnkitRai01
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the required permutation
static void findPerm(int []Q, int n)
{
int minval = 0, qsum = 0;
for (int i = 0; i < n - 1; i++)
{
// Each element in P' is like a
// cumulative sum in Q
qsum += Q[i];
// minval is the minimum
// value in P'
if (qsum < minval)
minval = qsum;
}
int []P = new int[n];
P[0] = 1 - minval;
// To check if each entry in P
// is from the range [1, n]
bool permFound = true;
for (int i = 0; i < n - 1; i++)
{
P[i + 1] = P[i] + Q[i];
// Invalid permutation
if (P[i + 1] > n || P[i + 1] < 1)
{
permFound = false;
break;
}
}
// If a valid permutation exists
if (permFound)
{
// Print the permutation
for (int i = 0; i < n; i++)
{
Console.Write(P[i]+ " ");
}
}
else
{
// No valid permutation
Console.Write(-1);
}
}
// Driver code
public static void Main(String[] args)
{
int []Q = { -2, 1 };
int n = 1 + Q.Length;
findPerm(Q, n);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript implementation of the approach
// Function to find the required permutation
function findPerm(Q, n)
{
var minval = 0, qsum = 0;
for(var i = 0; i < n - 1; i++)
{
// Each element in P' is like a
// cumulative sum in Q
qsum += Q[i];
// minval is the minimum
// value in P'
if (qsum < minval)
minval = qsum;
}
var P = Array(n);
P[0] = 1 - minval;
// To check if each entry in P
// is from the range [1, n]
var permFound = true;
for(var i = 0; i < n - 1; i++)
{
P[i + 1] = P[i] + Q[i];
// Invalid permutation
if (P[i + 1] > n || P[i + 1] < 1)
{
permFound = false;
break;
}
}
// If a valid permutation exists
if (permFound)
{
// Print the permutation
for(var i = 0; i < n; i++)
{
document.write( P[i] + " ");
}
}
else
{
// No valid permutation
document.write( -1);
}
}
// Driver code
var Q = [ -2, 1 ];
var n = 1 + Q.length;
findPerm(Q, n);
// This code is contributed by famously
</script>
Time Complexity: O(n)
Auxiliary Space: O(n)
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