Here is my solution most optimized in time. Hope it works fine with you all.
Intuition\nThe problem to me appears to first verify whether given array can be decomposed into such valid pairs, so at very first step i figure out patterns of
# Intuition\nThe problem to me appears to first verify whether given array can be decomposed into such valid pairs, so at very first step i figure out patterns of all valid and invalid arrays.\n\n# Approach\nAfter figuring out the pattern for valid array that holds a solution the next step was to convert any array into a valid one.\nHere the valid array to me seems any array which can be converted into either monotonically increasing or decreasing.\nThen for finding a solution to any such array it was observe that the prefix sum pattern was there by running through multiple test cases.\n\nNote- My first description of solution i wish i can give clear explanation about prefix sum pattern, hope this image works well.\n![monot.jpg](https://assets.leetcode.com/users/images/4c9cd86c-bbfa-4657-acc2-63157a7a10dc_1723671597.1515057.jpeg)\n\n# Complexity\n- Time complexity:\nThe solution runs in O(max*n)\nwhere max is maximum element in array and n is length of array\n\n- Space complexity:\nI maintained a space of O(n) to maintain the prefix sum.\n(this might be avoided not sure but need to dig more into it)\n\n# Code\n```\nclass Solution {\n    //to check whether its a non decreasing array if not whether its possible to convert it\n    //after transforming a array into non decreasing array it will return minimum element\n    \n    public int f1(int[] nums){\n        int la=nums[0];\n        int min=nums[0];\n        for(int i=1;i<nums.length;i++){\n            if(nums[i]<la){\n                if(la-nums[i]<=min){\n                    min=min-(la-nums[i]);\n                }\n                else return -1;\n            }\n            la=nums[i];\n        }\n        return min;\n    }\n    public int countOfPairs(int[] nums) {\n        int t=f1(nums);\n\n        if(t==-1)\n            return 0;  //no pair exist\n        long res=1;\n        if(t==0)\n            return 1;  //only {0,0,....,0,0} is one valid pair possible\n        int mod=1000000007;\n        long[] arr=new long[nums.length]; //to maintain prefix sum\n        Arrays.fill(arr,1);\n        for(int i=1;i<=t;i++){\n            long sum=0;\n            for(int j=nums.length-1;j>=0;j--){\n                sum+=arr[j];\n                arr[j]=sum%mod;\n            }\n            res=(res+sum)%mod;\n        }\n        return (int)res%mod;\n    }\n}\n```
['Java']