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# Most worth you possibly can get hold of at index 0 by performing given operation

Given an array nums[] of measurement N, Discover the most worth you possibly can obtain at index 0 after performing the operation the place in every operation improve the worth at index 0 by 1 and reduce the worth at index i by 1 such that nums[0] < nums[i] (1 ≤ i ≤ N-1). You may carry out as many strikes as you desire to (presumably, zero).

Examples:

Enter: nums[] = {1, 2, 3}
Output: 3
Rationalization: nums[0] < nums[1], Subsequently nums[0] + 1 and nums[1] – 1. Now,  nums[0] = 2, nums[0] < nums[2], once more repeat the step. Now nums[0] turns into 3 which is the utmost potential worth we will get at index 0.

Enter: nums[] = {1, 2, 2}
Output: 2

Method: The issue may be solved primarily based on the next thought:

In an effort to obtain most worth at index 0, type the array from index 1 to the final index and may begin iterating it from index 1 and if at any level nums[i] is discovered to be better than the factor current at index 0, in line with the given operation nums[0] get elevated and nums[i] get decreased. So, nums[0] get elevated by the quantity (nums[i] – nums[0] + 1) / 2 when encountered with a better worth.

Comply with the steps talked about under to implement the concept:

• Retailer the preliminary worth current at index 0 in a variable.
• Kind the array from index 1 to the final index.
• Iterate from index 1 and if discovered nums[i] > nums[0], do the given operation
• Worth at index 0 will get elevated by the (nums[i] – nums[0] +1) / 2
• Return the worth at index 0 as the required reply.

Under is the implementation of the above method:

## C++

 ` `  `#embody ` `utilizing` `namespace` `std;` ` `  `int` `maximumValue(vector<``int``>& nums, ``int` `n)` `{` `    ` `    ``int` `value_at_index_0 = nums[0];` ` `  `    ` `    ` `    ``type(nums.start() + 1, nums.finish());` ` `  `    ` `    ``for` `(``int` `i = 1; i < n; i++) {` ` `  `        ` `        ` `        ` `        ``if` `(nums[i] > value_at_index_0) {` ` `  `            ` `            ``value_at_index_0` `                ``+= (nums[i] - value_at_index_0 + 1) / 2;` `        ``}` `    ``}` ` `  `    ` `    ``return` `value_at_index_0;` `}` ` `  `int` `principal()` `{` `    ``vector<``int``> nums = { 1, 2, 3 };` `    ``int` `N = nums.measurement();` ` `  `    ` `    ``cout << maximumValue(nums, N);` ` `  `    ``return` `0;` `}`

Time Complexity: O(N * log N)
Auxiliary Area: O(1)

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