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# Minimum Operations To Make A Subsequence - Leetcode Solution

## Companies:

LeetCode: Minimum Operations To Make A Subsequence Leetcode Solution

Difficulty: Hard

Topics: greedy hash-table binary-search array

Problem Statement:

Given two integer arrays nums1 and nums2, return the length of the longest subsequence that appears in both arrays.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Constraints: 1 <= nums1.length, nums2.length <= 1000 0 <= nums1[i], nums2[i] <= 2000

Example:

Input: nums1 = [1,3,5,6,7], nums2 = [1,2,3,4,5,6]

Output: 4

Explanation:

The longest subsequence is [1,3,5,6] or [1,2,3,6] or [1,2,3,4,5,6], all of which have a length of 4.

Solution:

The given problem can be solved using dynamic programming. The subproblem can be defined as follows: Let dp[i][j] denote the length of the longest common subsequence that can be formed from the given two arrays up to the i-th element of nums1 and the j-th element of nums2.

The base cases are dp[0][j] = 0 and dp[i][0] = 0, as there cannot be any common subsequence between an empty array and any other array.

If nums1[i] == nums2[j], then dp[i][j] = 1 + dp[i-1][j-1], as the current element can be added to the longest common subsequence formed till the i-1th element of nums1 and j-1th element of nums2.

Otherwise, dp[i][j] = max(dp[i-1][j], dp[i][j-1]), as there can be two possibilities to form the longest common subsequence i.e., either by removing the i-th element of nums1 or by removing the j-th element of nums2.

At the end, dp[nums1.length][nums2.length] will give the length of the longest common subsequence that can be formed from the given two arrays.

Implementation:

public int minOperations(int[] nums1, int[] nums2) {

```
int n1 = nums1.length;
int n2 = nums2.length;
if (n1 * 6 < n2 || n2 * 6 < n1) {
return -1;
}
int[] freq1 = new int[7];
int[] freq2 = new int[7];
int diff = 0;
for (int num : nums1) {
freq1[num]++;
diff += num;
}
for (int num : nums2) {
freq2[num]++;
diff -= num;
}
if (diff > 0) {
freq1 = freq2;
freq2 = freq1;
diff = -diff;
}
int count = 0;
for (int i = 1; i <= 6; i++) {
int j = 7 - i;
while (freq1[i] > 0 && diff > 0 || freq2[j] > 0 && diff > 0) {
if (freq1[i] > 0) {
freq1[i]--;
diff -= 6 - i;
} else {
freq2[j]--;
diff -= j - 1;
}
count++;
}
}
return count;
```

}

## Minimum Operations To Make A Subsequence Solution Code

`1`