702903. Partition Point in the Array
Partition Point in the Array
Summary
Given an array of integers, find a partition point where the sum of elements on its left side is equal to the sum of elements on its right side. This problem involves arrays and data structures.
Detailed Explanation
To solve this problem, we can start by iterating over the array and calculating the cumulative sum at each step. We then compare the cumulative sum with the remaining sum (total sum minus the current cumulative sum). If they are equal, we have found the partition point.
Here's a step-by-step breakdown of the solution:
- Initialize variables:
cumulative_sumto store the running sum andtotal_sumto store the total sum of the array. - Iterate over the array:
- For each element, add it to
cumulative_sum. - Calculate
remaining_sumastotal_sum - cumulative_sum.
- For each element, add it to
- Compare
cumulative_sumwithremaining_sum. If they are equal, return the current index as the partition point.
The time complexity of this solution is O(n), where n is the size of the array, since we only need to iterate over the array once. The space complexity is O(1) because we only use a constant amount of space to store the cumulative and remaining sums.
Optimized Solutions
Java
public int findPartitionPoint(int[] arr) {
int total_sum = 0;
for (int num : arr) {
total_sum += num;
}
int cumulative_sum = 0;
for (int i = 0; i < arr.length; i++) {
cumulative_sum += arr[i];
if (cumulative_sum == (total_sum - cumulative_sum)) {
return i;
}
}
return -1; // partition point not found
}