Minimum Replacements to Sort the Array
Asked at Salesforce
Problem
Given an array of positive integers, you can replace any element with any of its divisors (smaller values that divide it evenly). Find the minimum number of replacements needed to make the array non-decreasing. The greedy strategy is to process from right to left, replacing each element with the smallest divisor that keeps it ≤ the next element.
Asked At
| Company | Difficulty | |
|---|---|---|
| Salesforce | HARD | View all Salesforce questions → |
How to Think About It
Brute force: try all possible replacements for each element and check if the array becomes sorted — exponential.
Think greedily from right to left: the last element should be as small as possible.
For each element arr[i], find the smallest value v such that v ≤ arr[i+1] and v divides arr[i].
The minimum replacements for arr[i] is ceil(arr[i] / arr[i+1]) - 1, using integer division.
Process from right to left, keeping track of the current maximum allowed value for each position.
Optimal Approach
Process the array from right to left. Maintain a variable representing the maximum allowed value for the current position. For each element arr[i], compute how many replacements are needed: the minimum value we can place is arr[i] divided by ceil(arr[i] / arr[i+1]). The number of replacements is ceil(arr[i] / arr[i+1]) - 1. Update the maximum allowed value for the next (leftward) element. Sum all replacements across the array. This greedy approach works because making elements as large as possible at each step minimizes the total replacements.
What Trips People Up in Real Interviews
Clarify that you can only replace an element with its divisors (not arbitrary smaller values).
Ask whether the array must be strictly increasing or non-decreasing (allows equal elements).
Explain the greedy insight: process from right to left to minimize the number of replacements.
Handle edge cases: single element arrays, already sorted arrays, and large values.
Discuss that the time complexity is O(n log M) where M is the max element value for finding divisors.
Solution Code
import math
def min_replacements(arr):
n = len(arr)
replacements = 0
for i in range(n - 2, -1, -1):
if arr[i] > arr[i + 1]:
k = math.ceil(arr[i] / arr[i + 1])
replacements += k - 1
arr[i] = arr[i] // k
return replacementsFrequently Asked Questions
What is the Minimum Replacements to Sort the Array problem?
Given an array of positive integers, you can replace any element with any of its divisors (smaller values that divide it evenly). Find the minimum number of replacements needed to make the array non-decreasing. The greedy strategy is to process from right to left, replacing each element with the smallest divisor that keeps it ≤ the next element.
How do you solve Minimum Replacements to Sort the Array?
The optimal approach is described in detail above, including step-by-step walkthroughs, complexity analysis, and solution code in Python. Scroll up to the "Optimal Approach" section.
What companies ask Minimum Replacements to Sort the Array?
Minimum Replacements to Sort the Array is asked at Salesforce. It is a hard difficulty problem.
What are common mistakes on Minimum Replacements to Sort the Array?
- Clarify that you can only replace an element with its divisors (not arbitrary smaller values).
- Ask whether the array must be strictly increasing or non-decreasing (allows equal elements).
- Explain the greedy insight: process from right to left to minimize the number of replacements.
- Handle edge cases: single element arrays, already sorted arrays, and large values.
- Discuss that the time complexity is O(n log M) where M is the max element value for finding divisors.