This problem asks for the minimum number of operations to make all elements in an array equal to a target value k. In one operation, you can choose any value currently in the array that is strictly greater than k and replace all occurrences of that value with a smaller value that is still greater than or equal to k. The challenge is to count the minimum steps required to bring everything down to k.
Companies like Lowe's and Bloomberg ask this to test basic array handling and the use of Hash Tables (or Sets). It's an introductory problem that assesses whether you can recognize that the "bulk replacement" rule means the number of operations depends only on the number of unique values in the array that are greater than k.
The pattern is Hash Table / Set usage. Since one operation replaces all occurrences of a value, we only care about the set of unique values present. If any value in the array is already smaller than k, it's impossible to reach k (since we can only replace values with smaller ones, but not below the replacement target). If all values are >= k, the answer is simply the count of unique values in the array that are strictly greater than k.
Array: [5, 2, 5, 4, 3], k = 2.
{2, 3, 4, 5}.{3, 4, 5}.The most common mistake is thinking you need to simulate the replacements. Another error is including the value k in the count of operations; if a number is already k, it doesn't need to be changed. Candidates also occasionally forget to check if there's any element already smaller than k, in which case they should immediately return -1.
Always pay attention to the phrase "all occurrences of that value." This is a huge hint that the total number of elements doesn't matter as much as the number of distinct elements. Sets are your best friend in these scenarios. Also, always check the "impossible" condition first to save time and simplify your logic.