explain median of two sorted array without code, use visual and animation
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We have two sorted arrays, Array A with 5 elements and Array B with 6 elements. Our goal is to find the median of all 11 elements without fully merging the arrays. Since we have 11 total elements, the median will be the 6th element when all are sorted together. We'll use a binary search approach to efficiently find the correct partition points.
Our strategy uses binary search on the smaller array. We choose a partition point i in Array A, then calculate the corresponding partition point j in Array B using the formula j equals m plus n plus 1 divided by 2, minus i. This ensures the left partitions contain exactly half the total elements. Watch as we try different partition points - the red lines show where we split each array.
To validate a partition, we check two conditions. First, the last element of A's left part must be less than or equal to the first element of B's right part. Second, the last element of B's left part must be less than or equal to the first element of A's right part. In this example with i equals 2 and j equals 4, we check if 3 is less than or equal to 21, which is true. But we also check if 19 is less than or equal to 8, which is false. Since one condition fails, this partition is invalid and we need to adjust using binary search.
Since our previous partition was invalid because 19 was greater than 8, we need to adjust our search. The binary search algorithm tells us to move the partition in Array A to the right to include more elements from A in the left partition. We change i from 2 to 3, which automatically changes j from 4 to 3. Now we check the new conditions: 8 is less than or equal to 19, which is true, but 11 is not less than or equal to 9, so this partition is still invalid. The binary search continues until we find the correct partition.
Today we'll explore how to find the median of two sorted arrays efficiently. Given two sorted arrays A and B, we want to find the median of their combined elements without actually merging the arrays. This is a classic problem that can be solved in logarithmic time using binary search.
The key insight is to use binary search to find the correct partition point. We need to partition both arrays such that the left half contains exactly half of the total elements, and all elements in the left half are less than or equal to all elements in the right half. This allows us to find the median efficiently.
The algorithm follows these key steps: First, we choose the smaller array for binary search to minimize time complexity. For each partition position i in array A, we calculate the corresponding position j in array B using the formula shown. We then check if the partition is valid by ensuring elements before the partition are smaller than elements after it. Based on these conditions, we adjust our search range until we find the correct partition.
Let's trace through the binary search process. We start with low equals 0 and high equals 5. We try partition i equals 2 and calculate j equals 4. We check our conditions: A at position 1 which is 3 is less than or equal to B at position 4 which is 21, so that's satisfied. However, B at position 3 which is 19 is not less than or equal to A at position 2 which is 8. This means we need to increase i, so we adjust our search range.
After the binary search process, we find the correct partition at i equals 3 and j equals 3. This creates a left half with 6 elements: 1, 3, 8 from Array A and 7, 11, 18 from Array B. The right half has 5 elements. Since we have 11 total elements, the median is the 6th element, which is the maximum of the left half. The maximum of 8 and 18 is 18, so our median is 18. This algorithm efficiently finds the median in logarithmic time without fully merging the arrays.