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Maximum Sequence Length | Collatz Conjecture

  • Difficulty Level : Easy
  • Last Updated : 26 Apr, 2022

Given an integer N. The task is to find the number in the range from 1 to N-1 which is having the maximum number of terms in its Collatz Sequence and the number of terms in the sequence.
The collatz sequence of a number N is defined as: 
 

  • If N is Odd then change N to 3*N + 1.
  • If N is Even then change N to N / 2.

For example let us have a look at the sequence when N = 13
13 -> 40 -> 20 -> 10 -> 5 > 16 -> 8 -> 4 -> 2 -> 1
Examples: 
 

Input: 10 
Output: (9, 20)
9 has 20 terms in its Collatz sequence
Input: 50 
Output: (27, 112) 
27 has 112 terms 
 

 

Approach: 
As in the above example discussed for N = 13, collatz sequence for N = 13 and N = 40 have similar terms except one, that ensures there may be an involvement of dynamic programming to store the answer for subproblems and reuse it.
But here normal memoization will not work because at one step we are either making a number large from itself ( in above example N = 13 is depending upon the solution of N = 40 ) or dividing by 2 ( N = 40 solution depends upon the solution of N = 20 ).
So instead of using a dp array we will use a Map/ dictionary data structure to store the solution of subproblems and will perform the normal operation as discussed in the collatz sequence.
Below is the implementation of the above approach:
 

C++




#include<bits/stdc++.h>
using namespace std;
 
int collatzLenUtil(int n, unordered_map<int,int>&collLenMap){
 
  // If value already
  // computed, return it
  if(collLenMap.find(n) != collLenMap.end())
    return collLenMap[n];
 
  // Base case
  if(n == 1)
    collLenMap[n] = 1;
 
  // Even case
  else if(n % 2 == 0){
    collLenMap[n] = 1 + collatzLenUtil(n/2 , collLenMap);
  }
 
  // Odd case
  else{
    collLenMap[n] = 1 + collatzLenUtil(3 * n + 1, collLenMap);
  }
 
  return collLenMap[n];
}
 
pair<int,int> collatzLen(int n){
 
  // Declare empty Map / Dict
  // to store collatz lengths
  unordered_map<int,int>collLenMap;
 
  collatzLenUtil(n, collLenMap);
 
  // Initialise ans and
  // its collatz length
  int num = -1, l = 0;
 
  for(int i=1;i<n;i++){
 
    // If value not already computed,
    // pass Dict to Helper function
    // and calculate and store value
    if(collLenMap.find(i)==collLenMap.end())
      collatzLenUtil(i, collLenMap);
 
    int cLen = collLenMap[i];
    if(l < cLen){
      l = cLen;
      num = i;
    }
 
  }
 
  // Return ans and
  // its collatz length
  return {num, l};
}
 
int main(){
 
  cout<<"("<<collatzLen(10).first<<", "<<collatzLen(10).second<<")"<<endl;
 
}
 
// This code is contributed by shinjanpatra

Python3




def collatzLenUtil(n, collLenMap):
     
    # If value already
    # computed, return it
    if n in collLenMap:
        return collLenMap[n]
     
    # Base case
    if(n == 1):
        collLenMap[n] = 1
 
    # Even case
    elif(n % 2 == 0):
        collLenMap[n] \
        = 1 \
           + collatzLenUtil(n//2, collLenMap)
 
    # Odd case
    else:
        collLenMap[n] \
        = 1 \
          + collatzLenUtil(3 * n + 1, collLenMap)
     
    return collLenMap[n]
 
def collatzLen(n):
     
    # Declare empty Map / Dict
    # to store collatz lengths
    collLenMap = {}
     
    collatzLenUtil(n, collLenMap)
 
    # Initialise ans and
    # its collatz length
    num, l =-1, 0
     
    for i in range(1, n):
         
        # If value not already computed,
        # pass Dict to Helper function
        # and calculate and store value
        if i not in collLenMap:
            collatzLenUtil(i, collLenMap)
         
        cLen = collLenMap[i]
        if l < cLen:
            l = cLen
            num = i
     
    # Return ans and
    # its collatz length
    return (num, l)
 
print(collatzLen(10))

Javascript




<script>
 
function collatzLenUtil(n, collLenMap){
     
    // If value already
    // computed, return it
    if(collLenMap.has(n))
        return collLenMap.get(n)
     
    // Base case
    if(n == 1)
        collLenMap.set(n, 1)
 
    // Even case
    else if(n % 2 == 0){
        collLenMap.set(n , 1 + collatzLenUtil(Math.floor(n/2) , collLenMap))
    }
 
    // Odd case
    else{
        collLenMap.set(n , 1 + collatzLenUtil(3 * n + 1, collLenMap))
    }
     
    return collLenMap.get(n);
}
 
function collatzLen(n){
     
    // Declare empty Map / Dict
    // to store collatz lengths
    let collLenMap = new Map()
     
    collatzLenUtil(n, collLenMap)
 
    // Initialise ans and
    // its collatz length
    let num = -1, l = 0
     
    for(let i=1;i<n;i++){
         
        // If value not already computed,
        // pass Dict to Helper function
        // and calculate and store value
        if(collLenMap.has(i)==false)
            collatzLenUtil(i, collLenMap)
         
        let cLen = collLenMap.get(i)
        if(l < cLen){
            l = cLen
            num = i
        }
 
    }
     
    // Return ans and
    // its collatz length
    return [num, l]
}
 
document.write(collatzLen(10))
 
// This code is contributed by shinjanpatra
 
</script>
Output: 
(9, 20)

 


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