# Partition given string in such manner that i’th substring is sum of (i-1)’th and (i-2)’th substring

Partition given string in such manner that i’th substring is sum of (i-1)’th and (i-2)’nd substring.

**Examples:**

Input :"11235813"Output :["1", "1", "2", "3", "5", "8", "13"]Input :"1111223"Output :["1", "11", "12", "23"]Input :"1111213"Output :["11", "1", "12", "13"]Input :"11121114"Output :[]

- Iterate through the given string by picking 3 numbers (first, seconds and third) at a time starting from one digit each.
- If first + second = third, recursively call check() with second as first and third as second. The third is picked based on next possible number of digits. (The result of addition of two numbers can have a max. of second’s & third’s digits + 1)
- Else, first increment the third (by adding more digits) till the limit (Here limit is sum of first and second).
- After incrementing third, following cases arise. a) When doesn’t match, increment the second offset. b) When doesn’t match, increment the first offset. c)
**Note:**Once a call to check() is made after incrementing the third offset, do not alter the second or first, as those are already finalized.

- When the end of the string is reached and the condition is satisfied, add “second” and “third” to the empty list. While rolling back the recursive stack, prepend the “first” to the list so the order is preserved.

**Implementation:**

## Java

`// Java program to check if we can partition a` `// string in a way that value of i-th string is` `// sum of (i-1)-th and (i-2)-th substrings.` `import` `java.util.LinkedList;` `public` `class` `SumArray {` `private` `static` `LinkedList<Integer> resultList =` ` ` `new` `LinkedList<>();` `private` `static` `boolean` `check(` `char` `[] chars, ` `int` `offset1,` ` ` `int` `offset2, ` `int` `offset3, ` `boolean` `freezeFirstAndSecond) {` ` ` `// Find subarrays according to given offsets` ` ` `int` `first = intOf(subArr(chars, ` `0` `, offset1));` ` ` `int` `second = intOf(subArr(chars, offset1, offset2));` ` ` `int` `third = intOf(subArr(chars, offset1 + offset2, offset3));` ` ` `// If condition is satisfied for current subarrays` ` ` `if` `(first + second == third) {` ` ` `// If whole array is covered.` ` ` `if` `(offset1 + offset2 + offset3 >= chars.length) {` ` ` `resultList.add(first);` ` ` `resultList.add(second);` ` ` `resultList.add(third);` ` ` `return` `true` `;` ` ` `}` ` ` ` ` `// Check if remaining array also satisfies the condition` ` ` `boolean` `result = check(subArr(chars, offset1,` ` ` `chars.length - offset1), offset2, offset3,` ` ` `Math.max(offset2, offset3), ` `true` `);` ` ` `if` `(result) {` ` ` `resultList.addFirst(first);` ` ` `}` ` ` `return` `result;` ` ` `}` ` ` `// If not satisfied, try incrementing third` ` ` `if` `(isValidOffSet(offset1, offset2, ` `1` `+ offset3, chars.length)) {` ` ` `if` `(check(chars, offset1, offset2, ` `1` `+ offset3, ` `false` `))` ` ` `return` `true` `; ` ` ` `}` ` ` `// If first and second have been finalized, do not` ` ` `// alter already computed results` ` ` `if` `(freezeFirstAndSecond)` ` ` `return` `false` `;` ` ` `// If first and second are not finalized` ` ` `if` `(isValidOffSet(offset1, ` `1` `+ offset2, Math.max(offset1,` ` ` `1` `+ offset2), chars.length)) {` ` ` `// Try incrementing second` ` ` `if` `(check(chars, offset1, ` `1` `+ offset2,` ` ` `Math.max(offset1, ` `1` `+ offset2), ` `false` `))` ` ` `return` `true` `; ` ` ` `}` ` ` `// Try incrementing first` ` ` `if` `(isValidOffSet(` `1` `+ offset1, offset2, Math.max(` `1` `+ offset1,` ` ` `offset2), chars.length)) {` ` ` `if` `(check(chars, ` `1` `+ offset1, offset2, Math.max(` `1` `+ offset1,` ` ` `offset2), ` `false` `))` ` ` `return` `true` `;` ` ` `}` ` ` `return` `false` `;` `}` `// Check if given three offsets are valid (Within array length` `// and third offset can represent sum of first two)` `private` `static` `boolean` `isValidOffSet(` `int` `offset1, ` `int` `offset2,` ` ` `int` `offset3, ` `int` `length) {` ` ` `return` `(offset1 + offset2 + offset3 <= length &&` ` ` `(offset3 == Math.max(offset1, offset2) ||` ` ` `offset3 == ` `1` `+ Math.max(offset1, offset2)));` `}` `// To get a subarray with starting from given` `// index and offset` `private` `static` `char` `[] subArr(` `char` `[] chars, ` `int` `index, ` `int` `offset) {` ` ` `int` `trueOffset = Math.min(chars.length - index, offset);` ` ` `char` `[] destArr = ` `new` `char` `[trueOffset];` ` ` `System.arraycopy(chars, index, destArr, ` `0` `, trueOffset);` ` ` `return` `destArr;` `}` `private` `static` `int` `intOf(` `char` `... chars) {` ` ` `return` `Integer.valueOf(` `new` `String(chars));` `}` `public` `static` `void` `main(String[] args) {` ` ` `String numStr = ` `"11235813"` `;` ` ` `char` `[] chars = numStr.toCharArray();` ` ` `System.out.println(check(chars, ` `1` `, ` `1` `, ` `1` `, ` `false` `));` ` ` `System.out.println(resultList);` `}` `}` |

**Output:**

true [1, 1, 2, 3, 5, 8, 13]

**Time Complexity:** **O(n)****Auxiliary Space: O(n)**