Find a pair from the given array with maximum nCr value in Python

PythonServer Side ProgrammingProgramming

Suppose we have an array arr with n integers, we have to find arr[i] and arr[j] from the array such that arr[i]Carr[j] is at large as possible. If there is more than one pair, return any one of them.

So, if the input is like [4, 1, 2], then the output will be 4 2 as 4C1 = 4, 4C2 = 6 and 2C1 = 2, so (4,2) is only pair as we want.

To solve this, we will follow these steps −

  • sort the list v
  • N := v[n - 1]
  • if N mod 2 is same as 1, then
    • first := N / 2 (integer division)
    • second := first + 1
    • left := -1, right := -1
    • temp := -1
    • for i in range 0 to n, do
      • if v[i] > first, then
        • temp := i
        • break
      • otherwise,
        • difference := first - v[i]
        • if difference < res1, then
          • res1 := difference
          • left := v[i]
    • right := v[temp]
    • difference1 := first - left
    • difference2 := right - second
    • if difference1 < difference2, then
      • print(N, left)
    • otherwise,
      • print(N, right)
  • otherwise,
    • max := N / 2 (integer division)
    • res := 3*(10^18)
    • R := -1
    • for i in range 0 to n - 1, do
      • difference := |v[i] - max|
      • if difference < res is non-zero, then
        • res := difference
        • R := v[i]
    • print(N, R)

Example

Let us see the following implementation to get better understanding −

 Live Demo

def findMatrixPair(v, n):
   v.sort()
   N = v[n - 1]
   if N % 2 == 1:
      first = N // 2
      second = first + 1
      res1, res2 = 3 * (10 ** 18), 3 * (10 ** 18)
      left, right = -1, -1
      temp = -1
      for i in range(0, n):
         if v[i] > first:
            temp = i
            break
         else:
            difference = first - v[i]
            if difference < res1:
               res1 = difference
               left = v[i]
      right = v[temp]
      difference1 = first - left
      difference2 = right - second
      if difference1 < difference2:
         print(N, left)
      else:
         print(N, right)
   else:
      max = N // 2
      res = 3 * (10 ** 18)
      R = -1
      for i in range(0, n - 1):
         difference = abs(v[i] - max)
         if difference < res:
         res = difference
         R = v[i]
      print(N, R)
v = [4,1,2]
n = len(v)
findMatrixPair(v, n)

Input

[4,1,2], 3

Output:

4 2
raja
Published on 27-Aug-2020 12:50:19
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