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# Maximize worth obtained in Array by leaping to the subsequent consecutive higher

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Given an array arr[] of dimension N, the duty is to search out the utmost worth that may be obtained by following the under circumstances:

• Choose any component (say okay) from the array and improve all different parts by 1.
• Within the subsequent step, soar solely to the index with a price okay+1.
• In the long run, you’re on the most worth.

Examples:

Enter: N = 4, arr[] ={1, 2, 1, 3}
Output: 3
Clarification: If began from index 0 with a peak of 1 unit, then,
the brand new worth of array will probably be [1, 3, 2, 4].
Then soar to the index with (1+1 = 2) ie 2nd index,
The up to date values are [2, 4, 2, 5]. Can’t be on the most worth at finish
The primary chosen worth was 3 at index 3.
The up to date values are [2, 3, 2, 3]. Max achieved -3. Therefore ans = 3;

Enter: N = 4, arr[]={1, 2, 3, 3}
Output: 4

Method: The issue may be solved based mostly on the next commentary:

On commentary, we are able to notice that for reaching most peak now we have two choices

• Straight deciding on the utmost heighted podium initially accessible.
• Selecting all the weather (say complete y) with similar worth (say x). So highest quantity that may be reached is (x + y – 1).

Comply with the under steps to resolve the issue:

• Type the array in growing order.
• Look for the span of the identical worth parts and get the utmost worth that may be achieved from that span utilizing the above thought.
• Carry out this for all accessible spans and retailer the utmost.
• Return the most because the required reply.

Under is the implementation of the above method.

## C++

 ` `  `#embody ` `utilizing` `namespace` `std;` ` `  `int` `maxVal(``int` `n, ``int` `a[])` `{` `    ` `    ` `    ``type(a, a + n);` `    ``int` `ans = 0, span = 0;` `    ``for` `(``int` `i = 1; i < n; i++) {` ` `  `        ` `        ``if` `(a[i - 1] == a[i]) {` `            ``span++;` `        ``}` `        ``else` `{` ` `  `            ` `            ` `            ``ans = max(ans, a[i - 1] + span);` `            ``span = 0;` `        ``}` `    ``}` `    ``ans = max(ans, a[n - 1] + span);` `    ``ans = max(ans, a[n - 1]);` ` `  `    ` `    ` `    ``return` `ans;` `}` ` `  `int` `important()` `{` `    ``int` `N = 4;` `    ``int` `arr[] = { 1, 2, 1, 3 };` ` `  `    ` `    ``cout << maxVal(N, arr) << endl;` `    ``return` `0;` `}`

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

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