Count of numbers from range [L, R] whose sum of digits is Y | Set 2
Last Updated :
26 Jul, 2025
Given three positive integers L, R and Y, the task is to count the numbers in the range [L, R] whose sum of digits is equal to Y
Examples:
Input: L = 500, R = 1000, Y = 6
Output: 3
Explanation:
Numbers in the range [500, 600] whose sum of digits is Y(= 6) are:
501 = 5 + 0 + 1 = 6
510 = 5 + 1 + 0 = 6
600 = 6 + 0 + 0 = 6
Therefore, the required output is 3.
Input: L = 20, R = 10000, Y = 14
Output: 540
Naive Approach: Refer to previous post to solve this problem by iterating over all the numbers in the range [L, R], and for every number, check if its sum of digits is equal to Y or not. If found to be true, then increment the count. Finally, print the count obtained.
Time Complexity: O(R - L + 1) * log10(R)
Auxiliary Space: O(1)
Efficient approach: To optimize the above approach, the idea is to use Digit DP using the following recurrence relation:
cntNum(i, sum, tight) = \sum^{9}_{i=0} cntNum(i + 1, (sum - i), tight & (i == end)
where, sum: Represents sum of digits.
tight: Check if sum of digits exceed Y or not.
end: Stores the maximum possible value of ith digit of a number.
cntNum(N, Y, tight): Returns the count of numbers in the range [0, X] whose sum of digits is Y.
Before moving into the DP solution, it is good practice to write down the recursive code.
Here is the recursive code -
C++
// C++ Program for the same approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of digits
// of numbers in the range [0, X]
int cntNum(string X, int i, int sum, int tight)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length() || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight != 0 ? X[i] - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0);
}
// Return res
return res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L,int R,int Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
string str = to_string(R);
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y,
1);
// Update str
str = to_string(L - 1);
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y,
1);
return (cntR - cntL);
}
// Driver code
int main()
{
int L = 20, R = 10000, Y = 14;
cout<<(UtilCntNumRange(L, R, Y));
}
// This code is contributed by shinjanpatra
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(String X, int i, int sum,
int tight)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length() || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight != 0 ? X.charAt(i) - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0);
}
// Return res
return res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L,int R,int Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
String str = String.valueOf(R);
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y,
1);
// Update str
str = String.valueOf(L - 1);
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y,
1);
return (cntR - cntL);
}
// Driver Code
public static void main (String[] args)
{
int L = 20, R = 10000, Y = 14;
System.out.print(UtilCntNumRange(L, R, Y));
}
}
// This code is contributed by Debojyoti Mandal
# Python program for the above approach
# Function to find the sum of digits
# of numbers in the range [0, X]
def cntNum(X, i, sum, tight):
# Check if count of digits in a number
# greater than count of digits in X
if (i >= len(X) or sum < 0):
# Check if sum of digits of a
# number is equal to Y
if (sum == 0):
return 1
return 0
# Stores count of numbers whose
# sum of digits is Y
res = 0
# Check if the number
# exceeds Y or not
end = ord(X[i]) - ord('0') if tight else 9
# Iterate over all possible
# values of i-th digits
for j in range(end+1):
# Update res
res += cntNum(X, i + 1, sum - j,1 if((tight > 0 and (j == end)) == True) else 0)
# Return res
return res
# Utility function to count the numbers in
# the range [L, R] whose sum of digits is Y
def UtilCntNumRange(L, R, Y):
# Base Case
if (R == 0 and Y == 0):
return 1
# Stores numbers in the form
# of its equivalent String
Str = str(R)
# Stores count of numbers
# in the range [0, R]
cntR = cntNum(Str, 0, Y,1)
# Update str
Str = str(L - 1)
# Stores count of numbers in
# the range [0, L - 1]
cntL = cntNum(Str, 0, Y, 1)
return (cntR - cntL)
# Driver Code
L, R, Y = 20, 10000, 14
print(UtilCntNumRange(L, R, Y))
# This code is contributed by shinjanpatra
// C# program for the above approach
using System;
class GFG
{
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(string X, int i, int sum, int tight)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.Length || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight != 0 ? X[i] - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++) {
// Update res
res += cntNum(
X, i + 1, sum - j,
(tight > 0 & (j == end)) == true ? 1 : 0);
}
// Return res
return res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L, int R, int Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
string str = R.ToString();
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y, 1);
// Update str
str = (L - 1).ToString();
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y, 1);
return (cntR - cntL);
}
// Driver Code
public static void Main(string[] args)
{
int L = 20, R = 10000, Y = 14;
Console.WriteLine(UtilCntNumRange(L, R, Y));
}
}
// This code is contributed by ukasp.
// JavaScript program for the above approach
// Function to find the sum of digits
// of numbers in the range [0, X]
function cntNum( X, i, sum, tight)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Stores count of numbers whose
// sum of digits is Y
var res = 0;
// Check if the number
// exceeds Y or not
var end = tight != 0 ? X[i].charCodeAt(0) - '0'.charCodeAt(0) : 9;
// Iterate over all possible
// values of i-th digits
for (var j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0);
}
// Return res
return res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
function UtilCntNumRange(L, R, Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
var str = (R).toString();
// Stores count of numbers
// in the range [0, R]
var cntR = cntNum(str, 0, Y,
1);
// Update str
str = (L - 1).toString();
// Stores count of numbers in
// the range [0, L - 1]
var cntL = cntNum(str, 0, Y,
1);
return (cntR - cntL);
}
// Driver Code
var L = 20, R = 10000, Y = 14;
document.write(UtilCntNumRange(L, R, Y));
// This code is contributed by shivanisinghss2110
Follow the steps below to solve the problem using DP.
- Initialize a 3D array dp[N][Y][tight] to compute and store the values of all subproblems of the above recurrence relation.
- Finally, return the value of dp[N][sum][tight].
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define M 1000
// Function to find the sum of digits
// of numbers in the range [0, X]
int cntNum(string X, int i, int sum,
int tight, int dp[M][M][2])
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length() || sum < 0) {
// If sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Check if current subproblem has
// already been computed
if (dp[sum][i][tight] != -1) {
return dp[sum][i][tight];
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight ? X[i] - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight & (j == end)), dp);
}
// Return res
return dp[sum][i][tight]=res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
int UtilCntNumRange(int L, int R, int Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent string
string str = to_string(R);
// Stores overlapping subproblems
int dp[M][M][2];
// Initialize dp[][][]
memset(dp, -1, sizeof(dp));
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y,
true, dp);
// Update str
str = to_string(L - 1);
// Initialize dp[][][]
memset(dp, -1, sizeof(dp));
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y,
true, dp);
return (cntR - cntL);
}
// Driver Code
int main()
{
int L = 20, R = 10000, Y = 14;
cout << UtilCntNumRange(L, R, Y);
}
// Java program for the above approach
import java.util.*;
class GFG{
static final int M = 1000;
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(String X, int i, int sum,
int tight, int dp[][][])
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length() || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Check if current subproblem has
// already been computed
if (dp[sum][i][tight] != -1) {
return dp[sum][i][tight];
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight != 0 ? X.charAt(i) - '0' : 9;
// Iterate over all possible
// values of i-th digits
for (int j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0, dp);
}
// Return res
return dp[sum][i][tight]=res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L, int R, int Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
String str = String.valueOf(R);
// Stores overlapping subproblems
int [][][]dp = new int[M][M][2];
// Initialize dp[][][]
for(int i = 0; i < M; i++)
{
for (int j = 0; j < M; j++) {
for (int k = 0; k < 2; k++)
dp[i][j][k] = -1;
}
}
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y,
1, dp);
// Update str
str = String.valueOf(L - 1);
// Initialize dp[][][]
for(int i = 0; i < M; i++)
{
for (int j = 0; j < M; j++) {
for (int k = 0; k < 2; k++)
dp[i][j][k] = -1;
}
}
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y,
1, dp);
return (cntR - cntL);
}
// Driver Code
public static void main(String[] args)
{
int L = 20, R = 10000, Y = 14;
System.out.print(UtilCntNumRange(L, R, Y));
}
}
// This code is contributed by shikhasingrajput
# Python program for the above approach
M = 1000
# Function to find the sum of digits
# of numbers in the range [0, X]
def cntNum(X, i, sum, tight, dp):
# Check if count of digits in a number
# greater than count of digits in X
if (i >= len(X) or sum < 0):
# Check if sum of digits of a
# number is equal to Y
if (sum == 0):
return 1
return 0
# Check if current subproblem has
# already been computed
if (dp[sum][i][tight] != -1):
return dp[sum][i][tight]
# Stores count of numbers whose
# sum of digits is Y
res, end = 0, 9
# Check if the number
# exceeds Y or not
if tight:
end = ord(X[i]) - ord('0')
# end = tight ? X[i] - '0' : 9;
# Iterate over all possible
# values of i-th digits
for j in range(end + 1):
# Update res
res += cntNum(X, i + 1, sum - j,
(tight & (j == end)), dp)
# Return res
dp[sum][i][tight] = res
return res
# Utility function to count the numbers in
# the range [L, R] whose sum of digits is Y
def UtilCntNumRange(L, R, Y):
# Base Case
if (R == 0 and Y == 0):
return 1
# Stores numbers in the form
# of its equivalent
strr = str(R)
# Stores overlapping subproblems
dp = [[[-1 for i in range(2)] for i in range(M)]
for i in range(M)]
# Initialize dp[][][]
# memset(dp, -1, sizeof(dp))
# Stores count of numbers
# in the range [0, R]
cntR = cntNum(strr, 0, Y, True, dp)
# Update str
strr = str(L - 1)
# Initialize dp[][][]
# memset(dp, -1, sizeof(dp))
# Stores count of numbers in
# the range [0, L - 1]
cntL = cntNum(strr, 0, Y, True, dp)
return (cntR - cntL)
# Driver Code
if __name__ == '__main__':
L, R, Y = 20, 10000, 14
print(UtilCntNumRange(L, R, Y))
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
class GFG{
static readonly int M = 1000;
// Function to find the sum of digits
// of numbers in the range [0, X]
static int cntNum(String X, int i, int sum,
int tight, int [,,]dp)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.Length || sum < 0)
{
// Check if sum of digits of a
// number is equal to Y
if (sum == 0)
{
return 1;
}
return 0;
}
// Check if current subproblem has
// already been computed
if (dp[sum, i, tight] != -1)
{
return dp[sum, i, tight];
}
// Stores count of numbers whose
// sum of digits is Y
int res = 0;
// Check if the number
// exceeds Y or not
int end = tight != 0 ? X[i] - '0' : 9;
// Iterate over all possible
// values of i-th digits
for(int j = 0; j <= end; j++)
{
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0, dp);
}
// Return res
return dp[sum][i][tight] = res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
static int UtilCntNumRange(int L, int R, int Y)
{
// Base Case
if (R == 0 && Y == 0)
{
return 1;
}
// Stores numbers in the form
// of its equivalent String
String str = String.Join("", R);
// Stores overlapping subproblems
int [,,]dp = new int[M, M, 2];
// Initialize [,]dp[]
for(int i = 0; i < M; i++)
{
for(int j = 0; j < M; j++)
{
for(int k = 0; k < 2; k++)
dp[i, j, k] = -1;
}
}
// Stores count of numbers
// in the range [0, R]
int cntR = cntNum(str, 0, Y,
1, dp);
// Update str
str = String.Join("",L - 1);
// Initialize [,]dp[]
for(int i = 0; i < M; i++)
{
for(int j = 0; j < M; j++)
{
for(int k = 0; k < 2; k++)
dp[i, j, k] = -1;
}
}
// Stores count of numbers in
// the range [0, L - 1]
int cntL = cntNum(str, 0, Y,
1, dp);
return (cntR - cntL);
}
// Driver Code
public static void Main(String[] args)
{
int L = 20, R = 10000, Y = 14;
Console.Write(UtilCntNumRange(L, R, Y));
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program for the above approach
let M = 1000;
// Function to find the sum of digits
// of numbers in the range [0, X]
function cntNum(X, i, sum, tight, dp)
{
// Check if count of digits in a number
// greater than count of digits in X
if (i >= X.length || sum < 0) {
// Check if sum of digits of a
// number is equal to Y
if (sum == 0) {
return 1;
}
return 0;
}
// Check if current subproblem has
// already been computed
if (dp[sum][i][tight] != -1) {
return dp[sum][i][tight];
}
// Stores count of numbers whose
// sum of digits is Y
let res = 0;
// Check if the number
// exceeds Y or not
let end = tight != 0 ? X[i].charCodeAt(0) - '0'.charCodeAt(0) : 9;
// Iterate over all possible
// values of i-th digits
for (let j = 0; j <= end; j++) {
// Update res
res += cntNum(X, i + 1, sum - j,
(tight > 0 & (j == end)) ==
true ? 1 : 0, dp);
}
// Return res
return dp[sum][i][tight]=res;
}
// Utility function to count the numbers in
// the range [L, R] whose sum of digits is Y
function UtilCntNumRange(L,R,Y)
{
// Base Case
if (R == 0 && Y == 0) {
return 1;
}
// Stores numbers in the form
// of its equivalent String
let str = (R).toString();
// Stores overlapping subproblems
let dp = new Array(M);
// Initialize dp[][][]
for(let i = 0; i < M; i++)
{
dp[i]=new Array(M);
for (let j = 0; j < M; j++) {
dp[i][j]=new Array(2);
for (let k = 0; k < 2; k++)
dp[i][j][k] = -1;
}
}
// Stores count of numbers
// in the range [0, R]
let cntR = cntNum(str, 0, Y,
1, dp);
// Update str
str = (L - 1).toString();
// Initialize dp[][][]
for(let i = 0; i < M; i++)
{
for (let j = 0; j < M; j++) {
for (let k = 0; k < 2; k++)
dp[i][j][k] = -1;
}
}
// Stores count of numbers in
// the range [0, L - 1]
let cntL = cntNum(str, 0, Y,
1, dp);
return (cntR - cntL);
}
// Driver Code
let L = 20, R = 10000, Y = 14;
document.write(UtilCntNumRange(L, R, Y));
// This code is contributed by patel2127
</script>
Time Complexity: O(Y * log10(R) * 10)
Auxiliary Space: O(Y * log10(R)
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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