Simple ,Effective and Intuitive Approach to Calculate Minimum Steps in Array Problem
Intuition\n Describe your first thoughts on how to solve this problem. \n figure out how many elements of size limit you need to reach a goal\n\n# Approach\n De
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n figure out how many elements of size limit you need to reach a goal\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1)Find the Total Difference:\nFirst, calculate the difference between your goal and the sum of your current elements. This tells you how much more (or less) you need to reach the goal.\n\n2)Determine Full Steps Needed:\nDivide this difference by limit to see how many full steps of size limit are required to cover that difference.\n\n3)Check for Any Leftover:\nIf there\u2019s no remainder when dividing (i.e., the difference is exactly divisible by limit), then you just need the number of full steps calculated.\n\nIf there\u2019s a remainder, you\u2019ll need one extra step to cover the leftover part.\n\nFor Example\nif you need to cover a total difference of, say, 11 units using steps of 5 units each:\nYou would need 2 full steps (10 units).\nSince 11 - 10 = 1 unit is left, you need one more step to cover this extra unit.\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nAccumulate function:O(N);\nArithmetic Operations:O(1);\n\noverall: O(N);\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int minElements(vector<int>& nums, int limit, int goal) {\n        long long sum=accumulate(nums.begin(),nums.end(),0LL);\n        if(sum==goal) return 0;\n        long long tempgoal;\n        tempgoal=goal-sum;\n        \n        int quot=abs(tempgoal)/limit;\n        if(abs(tempgoal)%limit==0) return quot;\n        else return quot+1;\n    }\n};\n```
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