It’s an easy problem with the description being:
You are given an integer array nums, an integer k, and an integer multiplier.
You need to perform k operations on nums. In each operation:
Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
Replace the selected minimum value x with x * multiplier.
Return an integer array denoting the final state of nums after performing all k operations.
Example 1:
Input: nums = [2,1,3,5,6], k = 5, multiplier = 2
Output: [8,4,6,5,6]
Explanation:
OperationResultAfter operation 1[2, 2, 3, 5, 6]After operation 2[4, 2, 3, 5, 6]After operation 3[4, 4, 3, 5, 6]After operation 4[4, 4, 6, 5, 6]After operation 5[8, 4, 6, 5, 6]
Example 2:
Input: nums = [1,2], k = 3, multiplier = 4
Output: [16,8]
Explanation:
OperationResultAfter operation 1[4, 2]After operation 2[4, 8]After operation 3[16, 8]
Constraints:
1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5]
Well, for this problem we can separate into three parts:
- Iterate k times
- Find the smallest number in the list
- Update smallest number with multiplier
Let’s also separate each into a method so it’s easier to understand:
class Solution {
public int[] getFinalState(int[] nums, int k, int multiplier) {
for(int i=0;i<k;i++) { // iterate k operations
final int smallestIndex = smallestNumberIndex(nums);
nums[smallestIndex] = applyMultiplier(nums[smallestIndex], multiplier);
}
return nums;
}
public int smallestNumberIndex(final int[] nums) {
int index = 0;
int x = nums[index];
for(int i=1;i<nums.length;i++) {
if(x > nums[i]) {
index = i;
x = nums[i];
}
}
return index;
}
public int applyMultiplier(final int x, final int multiplier) {
return x * multiplier;
}
}
Runtime: 1ms, faster than 66.48% of Java online submissions.
Memory Usage: 44.84 MB, less than 72.37% of Java online submissions.
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That’s it! If there is anything thing else to discuss feel free to drop a comment, if I missed anything let me know so I can update accordingly.
Until next post! :)