Table of Contents
LOGIC
Time Taken = 37 mins
- We have keep track of 2 number BIG NUMBER and SMALL NUMBER while iterating array from 2 to length-1.
- Rest of the code is simple.
- Most of the confusion occurs while initializing the index corresponding to
- bigNumberIndex and smallNumberIndex
- else condition was wrong at the beginning
- i.e. (nums[bigNumberIndex] > nums[i] ) which is going to be always true if we enter in else condition.
- Keep it simple, don’t write clever code.
JAVA CODE Maximum Product of Two Elements in an Array
class Solution {
public int maxProduct(int[] nums) {
if (nums.length == 2){
return ((nums[0]-1) * (nums[1]-1));
}
int smallNumberIndex = 0, bigNumberIndex = 1;
if (nums[0] > nums[1]){
bigNumberIndex = 0;
smallNumberIndex = 1;
}
for (int i = 2; i < nums.length; i++){
if (nums[i] > nums[bigNumberIndex]){
smallNumberIndex = bigNumberIndex;
bigNumberIndex = i;
} else if (nums[i] > nums[smallNumberIndex]) {
smallNumberIndex = i;
}
}
return ((nums[bigNumberIndex]-1) * (nums[smallNumberIndex]-1));
}
}Visit https: https://codeandalgo.com for more such contents
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