# Count of Numbers in Range where first digit is equal to last digit of the number

Given a range represented by two positive integers L and R. Find the count of numbers in the range where the first digit is equal to the last digit of the number.

**Examples:**

Input :L = 2, R = 60Output :13Explanation :Required numbers are 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44 and 55Input :L = 1, R = 1000Output :108

**Prerequisites **: Digit DP

There can be two approaches to solve this type of problem, one can be a combinatorial solution and other can be a dynamic programming based solution. Below is a detailed approach of solving this problem using a digit dynamic programming.

**Dynamic Programming Solution :** Firstly, if we are able to count the required numbers upto R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving for from zero to L – 1. Now, we need to define the DP states.

**DP States**:

- Since we can consider our number as a sequence of digits, one state is the
at which we are currently in. This position can have values from 0 to 18 if we are dealing with the numbers upto 10*position*^{18}. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9. - Second state is the
which defines the first digit of the number we are trying to build and can have values from 0 to 9.*firstD* - Third state is the
which defines the last digit of the number we are trying to build and can have values from 0 to 9.*lastD* - Another state is the boolean variable
which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, maximum limit of digit we can place is digit at current position in R.*tight*

In each recursive call, we set last digit as the digit we placed in the last position and we set first digit as the first non zero digit of the number. In the final recursive call, when we are at the last position if the first digit is equal to the last digit, return 1, otherwise 0.

Below is the implementation of the above approach.

## C++

`// CPP Program to find the count of ` `// numbers in a range where the number ` `// does not contain more than K non ` `// zero digits ` ` ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `const` `int` `M = 20; ` ` ` `// states - position, first digit, ` `// last digit, tight ` `int` `dp[M][M][M][2]; ` ` ` `// This function returns the count of ` `// required numbers from 0 to num ` `int` `count(` `int` `pos, ` `int` `firstD, ` `int` `lastD, ` ` ` `int` `tight, vector<` `int` `> num) ` `{ ` ` ` `// Last position ` ` ` `if` `(pos == num.size()) { ` ` ` ` ` `// If first digit is equal to ` ` ` `// last digit ` ` ` `if` `(firstD == lastD) ` ` ` `return` `1; ` ` ` `return` `0; ` ` ` `} ` ` ` ` ` `// If this result is already computed ` ` ` `// simply return it ` ` ` `if` `(dp[pos][firstD][lastD][tight] != -1) ` ` ` `return` `dp[pos][firstD][lastD][tight]; ` ` ` ` ` `int` `ans = 0; ` ` ` ` ` `// Maximum limit upto which we can place ` ` ` `// digit. If tight is 1, means number has ` ` ` `// already become smaller so we can place ` ` ` `// any digit, otherwise num[pos] ` ` ` `int` `limit = (tight ? 9 : num[pos]); ` ` ` ` ` `for` `(` `int` `dig = 0; dig <= limit; dig++) { ` ` ` `int` `currFirst = firstD; ` ` ` ` ` `// If the position is 0, current ` ` ` `// digit can be first digit ` ` ` `if` `(pos == 0) ` ` ` `currFirst = dig; ` ` ` ` ` `// In current call, if the first ` ` ` `// digit is zero and current digit ` ` ` `// is nonzero, update currFirst ` ` ` `if` `(!currFirst && dig) ` ` ` `currFirst = dig; ` ` ` ` ` `int` `currTight = tight; ` ` ` ` ` `// At this position, number becomes ` ` ` `// smaller ` ` ` `if` `(dig < num[pos]) ` ` ` `currTight = 1; ` ` ` ` ` `// Next recursive call, set last ` ` ` `// digit as dig ` ` ` `ans += count(pos + 1, currFirst, ` ` ` `dig, currTight, num); ` ` ` `} ` ` ` `return` `dp[pos][firstD][lastD][tight] = ans; ` `} ` ` ` `// This function converts a number into its ` `// digit vector and uses above function to compute ` `// the answer ` `int` `solve(` `int` `x) ` `{ ` ` ` `vector<` `int` `> num; ` ` ` `while` `(x) { ` ` ` `num.push_back(x % 10); ` ` ` `x /= 10; ` ` ` `} ` ` ` `reverse(num.begin(), num.end()); ` ` ` ` ` `// Initialize dp ` ` ` `memset` `(dp, -1, ` `sizeof` `(dp)); ` ` ` `return` `count(0, 0, 0, 0, num); ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `L = 2, R = 60; ` ` ` `cout << solve(R) - solve(L - 1) << endl; ` ` ` ` ` `L = 1, R = 1000; ` ` ` `cout << solve(R) - solve(L - 1) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find the count of ` `// numbers in a range where the number ` `// does not contain more than K non ` `// zero digits ` `import` `java.util.Collections; ` `import` `java.util.Vector; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `M = ` `20` `; ` ` ` ` ` `// states - position, first digit, ` ` ` `// last digit, tight ` ` ` `static` `int` `[][][][] dp = ` `new` `int` `[M][M][M][` `2` `]; ` ` ` ` ` `// This function returns the count of ` ` ` `// required numbers from 0 to num ` ` ` `static` `int` `count(` `int` `pos, ` `int` `firstD, ` ` ` `int` `lastD, ` `int` `tight, ` ` ` `Vector<Integer> num) ` ` ` `{ ` ` ` ` ` `// Last position ` ` ` `if` `(pos == num.size()) ` ` ` `{ ` ` ` ` ` `// If first digit is equal to ` ` ` `// last digit ` ` ` `if` `(firstD == lastD) ` ` ` `return` `1` `; ` ` ` `return` `0` `; ` ` ` `} ` ` ` ` ` `// If this result is already computed ` ` ` `// simply return it ` ` ` `if` `(dp[pos][firstD][lastD][tight] != -` `1` `) ` ` ` `return` `dp[pos][firstD][lastD][tight]; ` ` ` `int` `ans = ` `0` `; ` ` ` ` ` `// Maximum limit upto which we can place ` ` ` `// digit. If tight is 1, means number has ` ` ` `// already become smaller so we can place ` ` ` `// any digit, otherwise num[pos] ` ` ` `int` `limit = (tight == ` `1` `? ` `9` `: num.elementAt(pos)); ` ` ` ` ` `for` `(` `int` `dig = ` `0` `; dig <= limit; dig++) ` ` ` `{ ` ` ` `int` `currFirst = firstD; ` ` ` ` ` `// If the position is 0, current ` ` ` `// digit can be first digit ` ` ` `if` `(pos == ` `0` `) ` ` ` `currFirst = dig; ` ` ` ` ` `// In current call, if the first ` ` ` `// digit is zero and current digit ` ` ` `// is nonzero, update currFirst ` ` ` `if` `(currFirst == ` `0` `&& dig != ` `0` `) ` ` ` `currFirst = dig; ` ` ` ` ` `int` `currTight = tight; ` ` ` ` ` `// At this position, number becomes ` ` ` `// smaller ` ` ` `if` `(dig < num.elementAt(pos)) ` ` ` `currTight = ` `1` `; ` ` ` ` ` `// Next recursive call, set last ` ` ` `// digit as dig ` ` ` `ans += count(pos + ` `1` `, currFirst, ` ` ` `dig, currTight, num); ` ` ` `} ` ` ` `return` `dp[pos][firstD][lastD][tight] = ans; ` ` ` `} ` ` ` ` ` `// This function converts a number into its ` ` ` `// digit vector and uses above function to ` ` ` `// compute the answer ` ` ` `static` `int` `solve(` `int` `x) ` ` ` `{ ` ` ` `Vector<Integer> num = ` `new` `Vector<>(); ` ` ` `while` `(x > ` `0` `) ` ` ` `{ ` ` ` `num.add(x % ` `10` `); ` ` ` `x /= ` `10` `; ` ` ` `} ` ` ` ` ` `Collections.reverse(num); ` ` ` ` ` `// Initialize dp ` ` ` `for` `(` `int` `i = ` `0` `; i < M; i++) ` ` ` `for` `(` `int` `j = ` `0` `; j < M; j++) ` ` ` `for` `(` `int` `k = ` `0` `; k < M; k++) ` ` ` `for` `(` `int` `l = ` `0` `; l < ` `2` `; l++) ` ` ` `dp[i][j][k][l] = -` `1` `; ` ` ` ` ` `return` `count(` `0` `, ` `0` `, ` `0` `, ` `0` `, num); ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `L = ` `2` `, R = ` `60` `; ` ` ` `System.out.println(solve(R) - solve(L - ` `1` `)); ` ` ` ` ` `L = ` `1` `; ` ` ` `R = ` `1000` `; ` ` ` `System.out.println(solve(R) - solve(L - ` `1` `)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ` `// sanjeev2552 ` |

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**Output:**

13 108

**Time Complexity : **O(18 * 10 * 10 * 2 * 10), if we are dealing with the numbers upto 10^{18}

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