# Find the missing number in a string of numbers with no separator

Given a string consisting of some numbers, not separated by any separator. The numbers are positive integers and the sequence increases by one at each number except the missing number. The task is to find the missing number. The numbers will have no more than six digits. Print -1 if input sequence is not valid.

Examples:

```Input  : 89101113
Output : 12

Input  : 9899101102
Output : 100

Input  : 596597598600601602:
Output : 599

Input  : 909192939495969798100101
Output : 99

Input  : 11111211311411511
Output : -1
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to try all lengths from 1 to 6. For every length we try, we check if the current length satisfies the property of all consecutive numbers and one missing. An interesting thing is number of digits may change as we increment numbers. For example when we move to 100 from 99. To handle this situation, we find number of digits using log base 10.

Below is the implementation of above approach:

## C++

 `// C++ program to find a missing number in a ` `// string of consecutive numbers without any ` `// separator. ` `#include ` `using` `namespace` `std; ` `const` `int` `MAX_DIGITS = 6; ` ` `  `// gets the integer at position i with length m, ` `// returns it or -1, if none ` `int` `getValue(``const` `string& str, ``int` `i, ``int` `m) ` `{ ` `    ``if` `(i + m > str.length()) ` `        ``return` `-1; ` ` `  `    ``// Find value at index i and length m. ` `    ``int` `value = 0; ` `    ``for` `(``int` `j = 0; j < m; j++) ` `    ``{ ` `        ``int` `c = str[i + j] - ``'0'``; ` `        ``if` `(c < 0 || c > 9) ` `            ``return` `-1; ` `        ``value = value * 10 + c; ` `    ``} ` `    ``return` `value; ` `} ` ` `  `// Returns value of missing number ` `int` `findMissingNumber(``const` `string& str) ` `{ ` `    ``// Try all lengths for first number ` `    ``for` `(``int` `m=1; m<=MAX_DIGITS; ++m) ` `    ``{ ` `        ``// Get value of first number with current ` `        ``// length/ ` `        ``int` `n = getValue(str, 0, m); ` `        ``if` `(n == -1) ` `           ``break``; ` ` `  `        ``// To store missing number of current length ` `        ``int` `missingNo = -1; ` ` `  `        ``// To indicate whether the sequence failed ` `        ``// anywhere for current length. ` `        ``bool` `fail = ``false``; ` ` `  `        ``// Find subsequent numbers with previous number as n ` `        ``for` `(``int` `i=m; i!=str.length(); i += 1 + log10l(n)) ` `        ``{ ` `            ``// If we haven't yet found the missing number ` `            ``// for current length. Next number is n+2. Note ` `            ``// that we use Log10 as (n+2) may have more ` `            ``// length than n. ` `            ``if` `((missingNo == -1) && ` `                ``(getValue(str, i, 1+log10l(n+2)) == n+2)) ` `            ``{ ` `                ``missingNo = n + 1; ` `                ``n += 2; ` `            ``} ` ` `  `            ``// If next value is (n+1) ` `            ``else` `if` `(getValue(str, i, 1+log10l(n+1)) == n+1) ` `                ``n++; ` ` `  `            ``else` `            ``{ ` `                ``fail = ``true``; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``if` `(!fail) ` `          ``return` `missingNo; ` `    ``} ` `    ``return` `-1; ``// not found or no missing number ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``cout << findMissingNumber(``"99101102"``); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find a missing number in a ` `// string of consecutive numbers without any ` `// separator. ` ` `  `class` `GFG { ` ` `  `    ``static` `final` `int` `MAX_DIGITS = ``6``; ` ` `  `// gets the integer at position i with length m, ` `// returns it or -1, if none ` `    ``static` `int` `getValue(String str, ``int` `i, ``int` `m) { ` `        ``if` `(i + m > str.length()) { ` `            ``return` `-``1``; ` `        ``} ` ` `  `        ``// Find value at index i and length m. ` `        ``int` `value = ``0``; ` `        ``for` `(``int` `j = ``0``; j < m; j++) { ` `            ``int` `c = str.charAt(i + j) - ``'0'``; ` `            ``if` `(c < ``0` `|| c > ``9``) { ` `                ``return` `-``1``; ` `            ``} ` `            ``value = value * ``10` `+ c; ` `        ``} ` `        ``return` `value; ` `    ``} ` ` `  `// Returns value of missing number ` `    ``static` `int` `findMissingNumber(String str) { ` `        ``// Try all lengths for first number ` `        ``for` `(``int` `m = ``1``; m <= MAX_DIGITS; ++m) { ` `            ``// Get value of first number with current ` `            ``// length/ ` `            ``int` `n = getValue(str, ``0``, m); ` `            ``if` `(n == -``1``) { ` `                ``break``; ` `            ``} ` ` `  `            ``// To store missing number of current length ` `            ``int` `missingNo = -``1``; ` ` `  `            ``// To indicate whether the sequence failed ` `            ``// anywhere for current length. ` `            ``boolean` `fail = ``false``; ` ` `  `            ``// Find subsequent numbers with previous number as n ` `            ``for` `(``int` `i = m; i != str.length(); i += ``1` `+ Math.log10(n)) { ` `                ``// If we haven't yet found the missing number ` `                ``// for current length. Next number is n+2. Note ` `                ``// that we use Log10 as (n+2) may have more ` `                ``// length than n. ` `                ``if` `((missingNo == -``1``) ` `                        ``&& (getValue(str, i, (``int``) (``1` `+ Math.log10(n + ``2``))) == n + ``2``)) { ` `                    ``missingNo = n + ``1``; ` `                    ``n += ``2``; ` `                ``} ``// If next value is (n+1) ` `                ``else` `if` `(getValue(str, i, (``int``) (``1` `+ Math.log10(n + ``1``))) == n + ``1``) { ` `                    ``n++; ` `                ``} ``else` `{ ` `                    ``fail = ``true``; ` `                    ``break``; ` `                ``} ` `            ``} ` ` `  `            ``if` `(!fail) { ` `                ``return` `missingNo; ` `            ``} ` `        ``} ` `        ``return` `-``1``; ``// not found or no missing number ` `    ``} ` `// Driver code ` ` `  `    ``public` `static` `void` `main(String[] args) { ` `        ``System.out.println(findMissingNumber(``"99101102"``)); ` `    ``} ` `} ` `// This code is contributed by 29AjayKumar `

## C#

 `// C# program to find a missing number in a  ` `// string of consecutive numbers without any  ` `// separator.  ` `using` `System; ` `public` `class` `GFG {  ` ` `  `    ``static` `readonly` `int` `MAX_DIGITS = 6;  ` ` `  `// gets the integer at position i with length m,  ` `// returns it or -1, if none  ` `    ``static` `int` `getValue(String str, ``int` `i, ``int` `m) {  ` `        ``if` `(i + m > str.Length) {  ` `            ``return` `-1;  ` `        ``}  ` ` `  `        ``// Find value at index i and length m.  ` `        ``int` `value = 0;  ` `        ``for` `(``int` `j = 0; j < m; j++) {  ` `            ``int` `c = str[i + j] - ``'0'``;  ` `            ``if` `(c < 0 || c > 9) {  ` `                ``return` `-1;  ` `            ``}  ` `            ``value = value * 10 + c;  ` `        ``}  ` `        ``return` `value;  ` `    ``}  ` ` `  `// Returns value of missing number  ` `    ``static` `int` `findMissingNumber(String str) {  ` `        ``// Try all lengths for first number  ` `        ``for` `(``int` `m = 1; m <= MAX_DIGITS; ++m) {  ` `            ``// Get value of first number with current  ` `            ``// length/  ` `            ``int` `n = getValue(str, 0, m);  ` `            ``if` `(n == -1) {  ` `                ``break``;  ` `            ``}  ` ` `  `            ``// To store missing number of current length  ` `            ``int` `missingNo = -1;  ` ` `  `            ``// To indicate whether the sequence failed  ` `            ``// anywhere for current length.  ` `            ``bool` `fail = ``false``;  ` ` `  `            ``// Find subsequent numbers with previous number as n  ` `            ``for` `(``int` `i = m; i != str.Length; i += 1 + (``int``)Math.Log10(n)) {  ` `                ``// If we haven't yet found the missing number  ` `                ``// for current length. Next number is n+2. Note  ` `                ``// that we use Log10 as (n+2) may have more  ` `                ``// length than n.  ` `                ``if` `((missingNo == -1)  ` `                        ``&& (getValue(str, i, (``int``) (1 + Math.Log10(n + 2))) == n + 2)) {  ` `                    ``missingNo = n + 1;  ` `                    ``n += 2;  ` `                ``} ``// If next value is (n+1)  ` `                ``else` `if` `(getValue(str, i, (``int``) (1 + Math.Log10(n + 1))) == n + 1) {  ` `                    ``n++;  ` `                ``} ``else` `{  ` `                    ``fail = ``true``;  ` `                    ``break``;  ` `                ``}  ` `            ``}  ` ` `  `            ``if` `(!fail) {  ` `                ``return` `missingNo;  ` `            ``}  ` `        ``}  ` `        ``return` `-1; ``// not found or no missing number  ` `    ``}  ` `// Driver code  ` ` `  `    ``public` `static` `void` `Main() {  ` `        ``Console.WriteLine(findMissingNumber(``"99101102"``));  ` `    ``}  ` `}  ` `  ``//This code is contributed by PrinciRaj1992 `

Output:

```100
```

This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : 29AjayKumar, princiraj1992

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