✅ 🎯 📌 Simple Solution || Beats 100% in Time || Dynamic Programming ✅ 🎯 📌
Initialization:\n   - We start by initializing an array curr for the first element of nums.\n   - curr[i] stores the number of ways to form valid pairs where ar
1. **Initialization**:\n   - We start by initializing an array `curr` for the first element of `nums`.\n   - `curr[i]` stores the number of ways to form valid pairs where `arr1[0] = i` and `arr2[0] = nums[0] - i`. Each index in `curr` is initially set to `1` because any valid split of `nums[0]` has exactly one valid pair.\n\n2. **State Transition**:\n   - As we iterate through the elements of `nums`, we create a new array `newcurr` for each `num`.\n   - The loop that updates `newcurr` starts from the index `max(len(newcurr) - len(curr), 0)`. This ensures we don\'t access out-of-bounds indices when transitioning from `curr` to `newcurr`, especially when `newcurr` is larger than `curr`.\n   - For each possible value of `arr1[i]`, we use a cumulative sum (`val`) from the previous `curr` array to populate `newcurr`. This sum ensures that as `arr1` increases, the number of valid pairs correctly reflects the non-decreasing property of `arr1`.\n   - The cumulative sum is adjusted using `i + 1 - max(len(newcurr) - len(curr), 0)` to ensure correct indexing, especially when the sizes of `curr` and `newcurr` differ.\n\n3. **Final Calculation**:\n   - After processing all elements in `nums`, the sum of all elements in the final `curr` array gives the total count of valid pairs.\n   - Since the number can be very large, we take the result modulo \\(10^9 + 7\\).\n# Code\n```\nclass Solution:\n    def countOfPairs(self, nums: List[int]) -> int:\n        curr=[1 for i in range(nums[0]+1)]\n        for num in nums[1:]:\n            newcurr=[0 for i in range(num+1)]\n            val=curr[0]\n            for i in range(max(len(newcurr)-len(curr),0),len(newcurr)):\n                newcurr[i]+=val\n                if i+1-max(len(newcurr)-len(curr),0)<len(curr): val+=curr[i+1-max(len(newcurr)-len(curr),0)]\n            curr=newcurr\n        return sum(curr)%1_000_000_007\n```
['Dynamic Programming', 'Python3']