✅Easiest Basic C++ solution | Full Explanations with Dry Run | No advance Logiv✅
Very Easy solution..just imagine elements of array as I proceed\n\nSolution:\n\nclass Solution {\npublic:\n    int minOperations(vector<int>& nums) {\n        i
Very Easy solution..just imagine elements of array as I proceed\n\n**Solution:**\n```\nclass Solution {\npublic:\n    int minOperations(vector<int>& nums) {\n        int operations=0;  //cant keep 1 and do operations++ as for single ele array its inv\n        for(size_t i {1};i<nums.size();i++) //keep i=1 not 0 as if condition uses prev ele\n        {\n            if(nums[i-1]<nums[i])\n                continue;\n            else\n            {\n                //first calculate difference between previous and current term\n                int diff=nums[i-1]-nums[i];\n                diff++;\n                \n                //now add difference to the current element that is smaller than prev ele\n                nums[i]=nums[i]+diff;\n                \n                //increment count of operations by adding difference\n                operations+=diff;\n            }\n        }\n        return operations;\n        \n    }\n};\n```\n***Test:***\n\nLet\'s go through a dry run of the code using an example. Consider the input array nums = [1, 5, 2, 4, 1]\n\nIteration 1:\n- i = 1\n- nums[1 - 1] = nums[0] = 1\n- nums[1] = 5\n- Since 1 < 5, we continue.\n\nIteration 2:\n- i = 2\n- nums[2 - 1] = nums[1] = 5\n- nums[2] = 2\n- Since 5 >= 2, we proceed with the operation.\n  - Calculate the difference: diff = nums[1] - nums[2] + 1 = 5 - 2 + 1 = 4\n  - Increment nums[2] by diff: nums[2] += 4 => nums[2] = 6\n  - Increment operations by diff: operations += 4 => operations = 4\n\nIteration 3:\n- i = 3\n- nums[3 - 1] = nums[2] = 6\n- nums[3] = 4\n- Since 6 >= 4, we proceed with the operation.\n  - Calculate the difference: diff = nums[2] - nums[3] + 1 = 6 - 4 + 1 = 3\n  - Increment nums[3] by diff: nums[3] += 3 => nums[3] = 7\n  - Increment operations by diff: operations += 3 => operations = 7\n\nIteration 4:\n- i = 4\n- nums[4 - 1] = nums[3] = 7\n- nums[4] = 1\n- Since 7 >= 1, we proceed with the operation.\n  - Calculate the difference: diff = nums[3] - nums[4] + 1 = 7 - 1 + 1 = 7\n  - Increment nums[4] by diff: nums[4] += 7 => nums[4] = 8\n  - Increment operations by diff: operations += 7 => operations = 14\n\nEnd of iterations. Return operations = 14.\n
['Array', 'Greedy', 'C', 'C++']