<p>Welcome to the Editorial of <a href="https://tinyurl.com/IPCjunior-1-25-26"><b>IPC junior-1 (2025–26)</b></a>.</p>↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/A"><h2>A. Choco-Coco</h2></a>↵
<b>Author: </b>[user:Aditya_Dave,2025-10-16], [user:202401457,2025-10-16], [user:billu_1903,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
In this problem, it is given that Yashvi has <i>x</i> number of chocolates and Femina has <i>y</i> number of chocolates. The total number of children is <i>n</i>. To distribute the chocolates among the children, it is necessary for the total number of chocolates to be more than or equal to the number of children. Because by doing so, it is guaranteed that every child will have atleast one chocolate. So the answer is <b>YES</b> only if the sum of <i>x</i> and <i>y</i> is greater than or equal to number of children. Otherwise the answer is <b>NO</b>↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int x, y, n;↵
↵
// Input number of chocolates of type x and y↵
scanf("%d %d", &x, &y);↵
↵
// Input number of children↵
scanf("%d", &n);↵
↵
// Check if total chocolates are enough for all children↵
if ((x + y) >= n) {↵
printf("YES\n");↵
} else {↵
printf("NO\n");↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/B"><h2>B. Strongest Digit</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
The lock opens with the maximum digit of a given number. We extract each and every digit of the number by doing modulo 10 operation and then divide the number by 10 to discard the last digit. Then we take the maximum of all the digits and print the answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int t;↵
scanf("%d", &t); // taking the number of test cases↵
↵
while (t--) { // loop runs for each test case↵
long long n;↵
scanf("%lld", &n); // input number for the current test case↵
↵
int maxi = 0;↵
while (n > 0) {↵
int rem = n % 10; // extract last digit↵
if (rem > maxi) {↵
maxi = rem; // update maximum digit↵
}↵
n /= 10; // remove last digit↵
}↵
printf("%d\n", maxi); // print maximum digit↵
}↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/C1"><h2>C1. Contest Glitch (Easy Version)</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Array, Implementation, Brute Force </spoiler>↵
↵
<spoiler summary="Solution">↵
In this question, we are told to find the first non negative missing number in the array. To do this, as the constraints are very low, we can run a for loop from 0 to 100 and for every number, we can check whether it occurs in the array or not. If the number does not occur in the array, we finally print the number and move on. So, this problem requires the use of nested for loops.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int n;↵
scanf("%d", &n); // Read the size of the array↵
↵
int a[n]; // Declare an array of size n↵
↵
// Read the array elements↵
for (int i = 0; i < n; i++) {↵
scanf("%d", &a[i]);↵
}↵
↵
// Check for the smallest missing number in range 0 to 100↵
for (int i = 0; i <= 100; i++) {↵
int found = 0; // Flag to check if current number i exists in array↵
↵
// Search for current number i in the array↵
for (int j = 0; j < n; j++) {↵
if (a[j] == i) {↵
found = 1; // Number found in array↵
break; // No need to check further↵
}↵
}↵
↵
// If number i is not found in the array↵
if (!found) {↵
printf("%d\n", i); // Print the missing number↵
return 0; // Exit the program↵
}↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/C2"><h2>C2. Contest Glitch (Hard Version)</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Array, Implementation, Hashing </spoiler>↵
↵
<spoiler summary="Solution">↵
As in the continuation of the previous question, if we paste the same code here, we will get a time limit exceeded error, due to tough constraints. So, we need to optimize our code to tailor the needs of this question. We will maintain a hash array of size <i>n</i>+1 initialized to -1 for all elements, where <i>n</i> is the size of the array. While traversing in the array, if we encounter a number less than or equal to <i>n</i>, we mark its presence in the hash array, if the number is greater than <i>n</i>, then we discard it due to pigeonhole principle. Finally, we initialize our answer variable by <i>n</i>+1, and traverse through the hash array, if the value at any particular index is -1, then we print the value and not see ahead. This will be our final answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int n;↵
scanf("%d", &n);↵
↵
int a[n];↵
int present[100001] = {0}; // Array to mark which numbers exist↵
↵
// Read array and mark present numbers↵
for (int i = 0; i < n; i++) {↵
scanf("%d", &a[i]);↵
present[a[i]] = 1;↵
}↵
↵
// Find the smallest number not present in the array (MEX)↵
int mex = 0;↵
while (present[mex]) {↵
mex++;↵
}↵
↵
printf("%d\n", mex); // Print MEX↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/D"><h2>D. Can Mahek Cheat ?</h2></a>↵
<b>Author: </b>[user:Dynamo_14,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> 2D-Array, Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
<i>n</i> is the number of rows, <i>m</i> is the number of columns↵
In this problem, if we apply the general solution using a 2-D array, it will give memory limit exceeded error. To optimize, we first find out the row and column in which Mahek is seated. The row can be found out by the formula <i>i</i> = ((<i>s</i>-1) / <i>m</i>) + 1, and the column can be found out by the formula <i>j</i> = ((<i>s</i>-1) % <i>m</i>) + 1. Then, for each invigilator, we check whether Mahek's seat number is in their range or not by checking whether <i>i</i> >= <i>r1</i> and <i>i</i> <= <i>r2</i> and <i>j</i> >= <i>c1</i> and <i>j</i> <= <i>c2</i>. Similarly for the second invigilator. If any one of these conditions become true, we print <b>NO</b>, else we print <b>YES</b>.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int s, n, m;↵
↵
// Read Mahek's seat number, rows, and columns↵
scanf("%d %d %d", &s, &n, &m);↵
↵
// Calculate Mahek's seat position (row and column)↵
// Seat numbering is row-wise starting from 1↵
int mahek_row = (s — 1) / m + 1;↵
int mahek_col = (s — 1) % m + 1;↵
↵
// Read first invigilator's region↵
int r1, c1, r2, c2;↵
scanf("%d %d %d %d", &r1, &c1, &r2, &c2);↵
↵
// Read second invigilator's region ↵
int r3, c3, r4, c4;↵
scanf("%d %d %d %d", &r3, &c3, &r4, &c4);↵
↵
// Check if Mahek is in first invigilator's region↵
int in_first = (mahek_row >= r1 && mahek_row <= r2 && ↵
mahek_col >= c1 && mahek_col <= c2);↵
↵
// Check if Mahek is in second invigilator's region↵
int in_second = (mahek_row >= r3 && mahek_row <= r4 && ↵
mahek_col >= c3 && mahek_col <= c4);↵
↵
// If Mahek is NOT in either region, he can cheat safely↵
if (!in_first && !in_second) {↵
printf("Yes\n");↵
} else {↵
printf("No\n");↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/E"><h2>E. Minimal K-Sum</h2></a>↵
<b>Author: </b>[user:Raze07,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Hashing, Implementation, Prefix sum, Sliding Window </spoiler>↵
↵
<spoiler summary="Solution">↵
This problem can be done using the prefix sum kind of logic. In C++, it provides a functionality of a map, which stores in a key-value fashion. We start iterating from the 0<sup>th</sup> index taking sum of the array, for every index, we store the index of latest possible <i>sum % k</i> to find the minimum length of <i>r</i> — <i>l</i> + 1. Then, from the formula for prefix sum on subarrays, if (prefix[<i>r</i>]-prefix[<i>l</i>-1]) % <i>k</i> = 0, then prefix[<i>r</i>] % <i>k</i> = prefix[<i>l</i>-1] % <i>k</i>, and the entry for map[0] will be initialized with -1 to preserve the structure of the problem. Then, for every index, we check whether <i>sum % k</i> belong in the map or not, if it does belong, then we take the min length of every possible subarray and store it in the variable, which will give us the final answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-cpp">#include <bits/stdc++.h>↵
using namespace std;↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
int t;↵
cin >> t; // Number of test cases↵
↵
while (t--) {↵
int n;↵
long long k;↵
cin >> n >> k;↵
↵
vector<long long> a(n);↵
for (int i = 0; i < n; i++) {↵
cin >> a[i]; // Input array elements↵
}↵
↵
// Map to store the last seen index for each prefix sum remainder modulo k↵
map<long long, int> seen;↵
seen[0] = -1; // Initialize remainder 0 at index -1↵
↵
long long prefix = 0;↵
int ans = n + 1; // Initialize with a value larger than maximum possible↵
↵
for (int i = 0; i < n; i++) {↵
prefix = (prefix + a[i]) % k;↵
↵
// Normalize remainder to be in range [0, k-1]↵
long long rem = prefix;↵
if (rem < 0) rem += k;↵
↵
// Check if we've seen this remainder before↵
if (seen.find(rem) != seen.end()) {↵
int prev = seen[rem];↵
ans = min(ans, i - prev); // Update minimum subarray length↵
}↵
↵
// Update the last seen index for this remainder↵
seen[rem] = i;↵
}↵
↵
// Output result↵
if (ans == n + 1) {↵
cout << -1 << "\n"; // No valid subarray found↵
} else {↵
cout << ans << "\n"; // Minimum length of subarray divisible by k↵
}↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<p>Explanations of all solutions contributed by [user:Aditya_Dave,2025-10-16]. </p>↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/A"><h2>A. Choco-Coco</h2></a>↵
<b>Author: </b>[user:Aditya_Dave,2025-10-16], [user:202401457,2025-10-16], [user:billu_1903,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
In this problem, it is given that Yashvi has <i>x</i> number of chocolates and Femina has <i>y</i> number of chocolates. The total number of children is <i>n</i>. To distribute the chocolates among the children, it is necessary for the total number of chocolates to be more than or equal to the number of children. Because by doing so, it is guaranteed that every child will have atleast one chocolate. So the answer is <b>YES</b> only if the sum of <i>x</i> and <i>y</i> is greater than or equal to number of children. Otherwise the answer is <b>NO</b>↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int x, y, n;↵
↵
// Input number of chocolates of type x and y↵
scanf("%d %d", &x, &y);↵
↵
// Input number of children↵
scanf("%d", &n);↵
↵
// Check if total chocolates are enough for all children↵
if ((x + y) >= n) {↵
printf("YES\n");↵
} else {↵
printf("NO\n");↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/B"><h2>B. Strongest Digit</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
The lock opens with the maximum digit of a given number. We extract each and every digit of the number by doing modulo 10 operation and then divide the number by 10 to discard the last digit. Then we take the maximum of all the digits and print the answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int t;↵
scanf("%d", &t); // taking the number of test cases↵
↵
while (t--) { // loop runs for each test case↵
long long n;↵
scanf("%lld", &n); // input number for the current test case↵
↵
int maxi = 0;↵
while (n > 0) {↵
int rem = n % 10; // extract last digit↵
if (rem > maxi) {↵
maxi = rem; // update maximum digit↵
}↵
n /= 10; // remove last digit↵
}↵
printf("%d\n", maxi); // print maximum digit↵
}↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/C1"><h2>C1. Contest Glitch (Easy Version)</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Array, Implementation, Brute Force </spoiler>↵
↵
<spoiler summary="Solution">↵
In this question, we are told to find the first non negative missing number in the array. To do this, as the constraints are very low, we can run a for loop from 0 to 100 and for every number, we can check whether it occurs in the array or not. If the number does not occur in the array, we finally print the number and move on. So, this problem requires the use of nested for loops.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int n;↵
scanf("%d", &n); // Read the size of the array↵
↵
int a[n]; // Declare an array of size n↵
↵
// Read the array elements↵
for (int i = 0; i < n; i++) {↵
scanf("%d", &a[i]);↵
}↵
↵
// Check for the smallest missing number in range 0 to 100↵
for (int i = 0; i <= 100; i++) {↵
int found = 0; // Flag to check if current number i exists in array↵
↵
// Search for current number i in the array↵
for (int j = 0; j < n; j++) {↵
if (a[j] == i) {↵
found = 1; // Number found in array↵
break; // No need to check further↵
}↵
}↵
↵
// If number i is not found in the array↵
if (!found) {↵
printf("%d\n", i); // Print the missing number↵
return 0; // Exit the program↵
}↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/C2"><h2>C2. Contest Glitch (Hard Version)</h2></a>↵
<b>Author: </b>[user:Raj_Patel_7807,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Array, Implementation, Hashing </spoiler>↵
↵
<spoiler summary="Solution">↵
As in the continuation of the previous question, if we paste the same code here, we will get a time limit exceeded error, due to tough constraints. So, we need to optimize our code to tailor the needs of this question. We will maintain a hash array of size <i>n</i>+1 initialized to -1 for all elements, where <i>n</i> is the size of the array. While traversing in the array, if we encounter a number less than or equal to <i>n</i>, we mark its presence in the hash array, if the number is greater than <i>n</i>, then we discard it due to pigeonhole principle. Finally, we initialize our answer variable by <i>n</i>+1, and traverse through the hash array, if the value at any particular index is -1, then we print the value and not see ahead. This will be our final answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int n;↵
scanf("%d", &n);↵
↵
int a[n];↵
int present[100001] = {0}; // Array to mark which numbers exist↵
↵
// Read array and mark present numbers↵
for (int i = 0; i < n; i++) {↵
scanf("%d", &a[i]);↵
present[a[i]] = 1;↵
}↵
↵
// Find the smallest number not present in the array (MEX)↵
int mex = 0;↵
while (present[mex]) {↵
mex++;↵
}↵
↵
printf("%d\n", mex); // Print MEX↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/D"><h2>D. Can Mahek Cheat ?</h2></a>↵
<b>Author: </b>[user:Dynamo_14,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> 2D-Array, Math, Implementation </spoiler>↵
↵
<spoiler summary="Solution">↵
<i>n</i> is the number of rows, <i>m</i> is the number of columns↵
In this problem, if we apply the general solution using a 2-D array, it will give memory limit exceeded error. To optimize, we first find out the row and column in which Mahek is seated. The row can be found out by the formula <i>i</i> = ((<i>s</i>-1) / <i>m</i>) + 1, and the column can be found out by the formula <i>j</i> = ((<i>s</i>-1) % <i>m</i>) + 1. Then, for each invigilator, we check whether Mahek's seat number is in their range or not by checking whether <i>i</i> >= <i>r1</i> and <i>i</i> <= <i>r2</i> and <i>j</i> >= <i>c1</i> and <i>j</i> <= <i>c2</i>. Similarly for the second invigilator. If any one of these conditions become true, we print <b>NO</b>, else we print <b>YES</b>.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-c">#include <stdio.h>↵
↵
int main() {↵
int s, n, m;↵
↵
// Read Mahek's seat number, rows, and columns↵
scanf("%d %d %d", &s, &n, &m);↵
↵
// Calculate Mahek's seat position (row and column)↵
// Seat numbering is row-wise starting from 1↵
int mahek_row = (s — 1) / m + 1;↵
int mahek_col = (s — 1) % m + 1;↵
↵
// Read first invigilator's region↵
int r1, c1, r2, c2;↵
scanf("%d %d %d %d", &r1, &c1, &r2, &c2);↵
↵
// Read second invigilator's region ↵
int r3, c3, r4, c4;↵
scanf("%d %d %d %d", &r3, &c3, &r4, &c4);↵
↵
// Check if Mahek is in first invigilator's region↵
int in_first = (mahek_row >= r1 && mahek_row <= r2 && ↵
mahek_col >= c1 && mahek_col <= c2);↵
↵
// Check if Mahek is in second invigilator's region↵
int in_second = (mahek_row >= r3 && mahek_row <= r4 && ↵
mahek_col >= c3 && mahek_col <= c4);↵
↵
// If Mahek is NOT in either region, he can cheat safely↵
if (!in_first && !in_second) {↵
printf("Yes\n");↵
} else {↵
printf("No\n");↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<a href="https://mirror.codeforces.com/gym/641211/problem/E"><h2>E. Minimal K-Sum</h2></a>↵
<b>Author: </b>[user:Raze07,2025-10-16]↵
↵
<spoiler summary="Problem Tags"> Math, Hashing, Implementation, Prefix sum, Sliding Window </spoiler>↵
↵
<spoiler summary="Solution">↵
This problem can be done using the prefix sum kind of logic. In C++, it provides a functionality of a map, which stores in a key-value fashion. We start iterating from the 0<sup>th</sup> index taking sum of the array, for every index, we store the index of latest possible <i>sum % k</i> to find the minimum length of <i>r</i> — <i>l</i> + 1. Then, from the formula for prefix sum on subarrays, if (prefix[<i>r</i>]-prefix[<i>l</i>-1]) % <i>k</i> = 0, then prefix[<i>r</i>] % <i>k</i> = prefix[<i>l</i>-1] % <i>k</i>, and the entry for map[0] will be initialized with -1 to preserve the structure of the problem. Then, for every index, we check whether <i>sum % k</i> belong in the map or not, if it does belong, then we take the min length of every possible subarray and store it in the variable, which will give us the final answer.↵
</spoiler>↵
↵
<spoiler summary="Code"><pre><code class="language-cpp">#include <bits/stdc++.h>↵
using namespace std;↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
int t;↵
cin >> t; // Number of test cases↵
↵
while (t--) {↵
int n;↵
long long k;↵
cin >> n >> k;↵
↵
vector<long long> a(n);↵
for (int i = 0; i < n; i++) {↵
cin >> a[i]; // Input array elements↵
}↵
↵
// Map to store the last seen index for each prefix sum remainder modulo k↵
map<long long, int> seen;↵
seen[0] = -1; // Initialize remainder 0 at index -1↵
↵
long long prefix = 0;↵
int ans = n + 1; // Initialize with a value larger than maximum possible↵
↵
for (int i = 0; i < n; i++) {↵
prefix = (prefix + a[i]) % k;↵
↵
// Normalize remainder to be in range [0, k-1]↵
long long rem = prefix;↵
if (rem < 0) rem += k;↵
↵
// Check if we've seen this remainder before↵
if (seen.find(rem) != seen.end()) {↵
int prev = seen[rem];↵
ans = min(ans, i - prev); // Update minimum subarray length↵
}↵
↵
// Update the last seen index for this remainder↵
seen[rem] = i;↵
}↵
↵
// Output result↵
if (ans == n + 1) {↵
cout << -1 << "\n"; // No valid subarray found↵
} else {↵
cout << ans << "\n"; // Minimum length of subarray divisible by k↵
}↵
}↵
↵
return 0;↵
}↵
</code></pre></spoiler>↵
↵
↵
<p>Explanations of all solutions contributed by [user:Aditya_Dave,2025-10-16]. </p>↵
