Count Permutations That Are First Decreasing Then Increasing in C++

We are a variable num. The goal is to find the count of permutations of numbers between [1,num] in which numbers are first decreasing then increasing. For example if num=3 then numbers are 1,2,3. The permutations will be [ 3,1,2 ] and [2,1,3] and count is 2.

We know that in every permutation the change from decreasing of numbers to increasing of numbers will be decided based on position of 1 which is smallest. After each 1 the numbers will start increasing. For a permutation to decrease and then increase, 1 should lie between position 2 and num-1. [ → ...1…. → ].

If 1 is at the start then series will be fully increasing [ 1.. → ] , if it is at the end then series will be fully decreasing [ … → 1 ].

Let’s say we have num=4 then

Placing 1 at 2nd position, [ - , 1, - , - ]. For 1st position we can choose from ( 2,3,4) let’s say we pick 2, then the sequence will be [ 2,1,3,4]. So 3C1 permutations are possible in this case.

Placing 1 at 3rd position, [ -, -, 1, - ]. For 1st and 2nd positions select any two out of three (2,3,4). Total permutations will be ³C₂.

So total permutations will be = ³C₁ + ³C₂ for num=4

For any num x, count will be = ^x-1C₁ + ^x-1C₂+.....+^x-1C_c-2 = 2_x-1 - 2 from binomial theorem.

Let us understand with examples

Input − num=4

Output − Count of permutations that are first decreasing then increasing are: 6

Explanation − Permutations will be −

[ 2, 1, 3, 4 ], [ 3, 1, 2, 4 ], [ 4, 1, 2, 3 ] → 1 at 2nd position
[ 2, 3, 1, 4 ], [ 2, 4, 1, 3 ], [ 3, 4, 1, 2 ] → 1 at 3rd position

Input − num=6

Output − Count of permutations that are first decreasing then increasing are − 30

Explanation − Some Permutations will be −

[ 2, 1, 3, 4, 5, 6 ], [ 3, 1, 2, 4, 5, 6 ], [ 4, 1, 2, 3, 5, 6 ], [ 5, 1, 2, 3, 4, 6 ], [ 6, 1, 2, 3, 4, 5 ]
……[ 6, 5, 4, 3, 1, 2].

Approach used in the below program is as follows

In this approach we will make use of a binomial theorem to directly calculate the permutations from the above formula. Also we will create a function value(long long i, long long num) which returns i^num

• Take variable num as input.

• Function permutations_increase_decrease(int num) takes num and returns the count of permutations that are first decreasing then increasing from numbers 1 to num.

• Function value(long long i, long long num) is used to calculate ( inum) % temp. Where temp=1000000007.

• Inside permutations_increase_decrease(int num) take temp=1000000007.

• If num is 1 no permutation possible so return 0.

• Else set count = (value(2, num - 1) - 2) % temp ); using formula.

• Return count as result.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
long long value(long long i, long long num){
   int temp = 1000000007;
   if (num == 0){
      return 1;
   }
   long long j = value(i, num / 2) % temp;
   j = (j * j) % temp;
   if(num & 1){
      j = (j * i) % temp;
   }
   return j;
}
int permutations_increase_decrease(int num){
   int temp = 1000000007;
   if (num == 1){
      return 0;
   }
   int count = (value(2, num - 1) - 2) % temp;
   return count;
}
int main(){
   int num = 4;
   cout<<"Count of permutations that are first decreasing then increasing are: "<<permutations_increase_decrease(num);
   return 0;
}

Output

If we run the above code it will generate the following output −

Count of permutations that are first decreasing then increasing are: 6