Java Solution || 100% Optimized || 100% faster  || O(1) Space || O(N) time
class Solution {\n    public int minOperations(int[] nums) {\n        \n        int count =0;  // Operations Count\n        int nextVal  = nums[0]+1;   // We ch
class Solution {\n    public int minOperations(int[] nums) {\n        \n        int count =0;  // Operations Count\n        int nextVal  = nums[0]+1;   // We choose the next value to be our successive element\n        for(int i=1;i<nums.length;i++){\n            \n            if(nums[i]<=nums[i-1]){\n                count += nextVal-nums[i];   // No. of operations to change the present value to the Next Succesor\n                nums[i] = nextVal;  // change array value\n            }\n            else{\n                nextVal = Math.max(nextVal,nums[i]); // Change the successor to the max value that can be put\n            }\n            nextVal++;  // for the array to be strictly increasing, always inc the nextVal ;\n        }\n        return count;\n    }\n}
['Java']