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# Reduce operations to make String palindrome by incrementing prefix by 1

Given a string S of numbers of size N, the duty is to search out the minimal variety of operations required to alter a string into palindrome and we are able to carry out the next process on the string :

• Select an index i (0 ≤ i < N) and for all 0 ≤ j ≤ i, set Sj = Sj + 1 (i.e. add 1 to each aspect within the prefix of size i).

Be aware: Whether it is not possible to transform S to a palindrome, print −1.

Examples:

Enter: S = “1234”
Output: 3
Clarification: We are able to carry out the next operations:
Choose i=1, “1234”→”2234″
Choose i=2, “2234”→”3334″
Choose i=1, “3334”→”4334″
Therefore 3 variety of operations required to alter a string into palindrome.

Enter:  S = “2442”
Output: 0
Clarification: The string is already a palindrome.

Strategy: The issue could be solved primarily based on the next statement:

Initially examine if the string is already a palindrome or not. If it isn’t a palindrome, then it may be transformed right into a palindrome provided that all the weather in the best half of the string is larger than those within the left half and the distinction between the characters nearer to finish is larger or equal to the distinction between those nearer to the centre.

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

• First set i = 0, j = N -1 and max = IntegerMaximumValue and ans = 0.
• After that iterate a loop till j > i
• Verify if S[j] < S[i], whether it is true then we are able to’t change the string into palindrome and return -1.
• In any other case, take absolutely the distinction of S[j] and S[i] and evaluate it with ans to search out the utmost between them:
• If the utmost worth is lower than absolutely the distinction of S[j] and S[i], return -1.
• In any other case, max is absolutely the distinction between S[j] and S[i]
• Return ans which is the minimal variety of operations required to alter a string right into a palindrome.

Beneath is the implementation of the above strategy.

## Java

 ` `  `import` `java.io.*;` `import` `java.util.*;` ` `  `class` `GFG {` ` `  `    ` `    ` `    ``public` `static` `int` `minOperation(String str, ``int` `n)` `    ``{` `        ``int` `j = n - ``1``;` `        ``int` `i = ``0``;` `        ``int` `max = Integer.MAX_VALUE;` `        ``int` `ans = ``0``;` `        ``whereas` `(j > i) {` `            ``if` `(str.charAt(j) < str.charAt(i)) {` `                ``return` `-``1``;` `            ``}` `            ``int` `ok = Math.abs(str.charAt(j) - str.charAt(i));` `            ``ans = Math.max(ans, ok);` `            ``if` `(max < ok) {` `                ``return` `-``1``;` `            ``}` `            ``max = ok;` `            ``i++;` `            ``j--;` `        ``}` `        ``return` `ans;` `    ``}` ` `  `    ` `    ``public` `static` `void` `primary(String[] args)` `    ``{` `        ``String S = ``"1234"``;` `        ``int` `N = S.size();` ` `  `        ` `        ``System.out.println(minOperation(S, N));` `    ``}` `}`

Time Complexity: O(N)
Auxiliary House: O(1)

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