Given an array a[], integer K and an integer X (which is initially initialized to 0). Our task is to find the minimum number of moves required to update the array such that each of its elements is divisible by K by performing the following operations:
- Choose one index i from 1 to N and increase ai by X and then increase X by 1. This operation cannot be applied more than once to each element of the array
- Only increase the value of X by 1.
Examples:
Input: K = 3, a = [1, 2, 2, 18]
Output: 5
Explanation:
Initially X = 0 hence update X to 1.
For X = 1 add X to the second element of array to make the array [1, 3, 2, 18] and increase X by 1.
For X = 2 add X to the first element of array [3, 3, 2, 18] and increase X by 1.
For X = 3 just increase X by 1.
For X = 4 add X to the third element of array to make the array [3, 3, 6, 18] and increase X by 1.
At last, the array becomes [3, 3, 6, 18] where all the elements are divisible by K = 3.
Input: K = 5, a[] = [15, 25, 5, 10, 20, 1005, 70, 80, 90, 100]
Output: 0
Explanation:
Here all elements are already divisible by 5.
Approach: The main idea is to find the maximum value of X that is needed to update the elements of the array to make it divisible by K.
- To do this we need to find the maximum value of (K - (ai mod K)) to add to the array elements to make it divisible by K.
- However, there can be equal elements so keep track of the number of such elements, using map data structures.
- When found another such element in the array then update the answer with (K - (ai mod K)) + (K * number of Equal Elements) because with every move we increase X by 1.
Below is the implementation of the above approach:
// C++ implementation to find the
// Minimum number of moves required to
// update the array such that each of
// its element is divisible by K
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// Minimum number of moves required to
// update the array such that each of
// its element is divisible by K
void compute(int a[], int N, int K)
{
// Initialize Map data structure
map<long, long> eqVal;
long maxX = 0;
// Iterate for all the elements
// of the array
for (int i = 0; i < N; i++) {
// Calculate the
// value to be added
long val = a[i] % K;
val = (val == 0 ? 0 : K - val);
// Check if the value equals
// to 0 then simply continue
if (val == 0)
continue;
// Check if the value to be
// added is present in the map
if (eqVal.find(val) != eqVal.end()) {
long numVal = eqVal[val];
// Update the answer
maxX = max(maxX,
val + (K * numVal));
eqVal[val]++;
}
else {
eqVal[val]++;
maxX = max(maxX, val);
}
}
// Print the required result
// We need to add 1 to maxX
// because we cant ignore the
// first move where initially X=0
// and we need to increase it by 1
// to make some changes in array
cout << (maxX == 0 ? 0 : maxX + 1)
<< endl;
}
// Driver code
int main()
{
int K = 3;
int a[] = { 1, 2, 2, 18 };
int N = sizeof(a) / sizeof(a[0]);
compute(a, N, K);
return 0;
}
// Java implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
import java.util.*;
class GFG{
// Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
static void compute(int a[], int N, int K)
{
// Initialize Map data structure
Map<Long,
Long> eqVal = new HashMap<Long,
Long>();
long maxX = 0;
// Iterate for all the elements
// of the array
for(int i = 0; i < N; i++)
{
// Calculate the
// value to be added
long val = a[i] % K;
val = (val == 0 ? 0 : K - val);
// Check if the value equals
// to 0 then simply continue
if (val == 0)
continue;
// Check if the value to be
// added is present in the map
if (eqVal.containsKey(val))
{
long numVal = eqVal.get(val);
// Update the answer
maxX = Math.max(maxX,
val + (K * numVal));
eqVal.put(val,
eqVal.getOrDefault(val, 0l) + 1l);
}
else
{
eqVal.put(val, 1l);
maxX = Math.max(maxX, val);
}
}
// Print the required result
// We need to add 1 to maxX
// because we cant ignore the
// first move where initially X=0
// and we need to increase it by 1
// to make some changes in array
System.out.println(maxX == 0 ? 0 : maxX + 1);
}
// Driver code
public static void main(String[] args)
{
int K = 3;
int a[] = { 1, 2, 2, 18 };
int N = a.length;
compute(a, N, K);
}
}
// This code is contributed by offbeat
# Python3 implementation to find the
# Minimum number of moves required to
# update the array such that each of
# its element is divisible by K
from collections import defaultdict
# Function to find the Minimum number
# of moves required to update the
# array such that each of its
# element is divisible by K
def compute(a, N, K):
# Initialize Map data structure
eqVal = defaultdict(int)
maxX = 0
# Iterate for all the elements
# of the array
for i in range(N):
# Calculate the
# value to be added
val = a[i] % K
if (val != 0):
val = K - val
# Check if the value equals
# to 0 then simply continue
if (val == 0):
continue
# Check if the value to be
# added is present in the map
if (val in eqVal):
numVal = eqVal[val]
# Update the answer
maxX = max(maxX,
val + (K * numVal))
eqVal[val] += 1
else:
eqVal[val] += 1
maxX = max(maxX, val)
# Print the required result
# We need to add 1 to maxX
# because we cant ignore the
# first move where initially X=0
# and we need to increase it by 1
# to make some changes in array
if maxX == 0:
print(0)
else:
print(maxX + 1)
# Driver code
if __name__ == "__main__":
K = 3
a = [ 1, 2, 2, 18 ]
N = len(a)
compute(a, N, K)
# This code is contributed by chitranayal
// C# implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
static void compute(int []a, int N, int K)
{
// Initialize Map data structure
Dictionary<long,
long> eqVal = new Dictionary<long,
long>();
long maxX = 0;
// Iterate for all the elements
// of the array
for(int i = 0; i < N; i++)
{
// Calculate the
// value to be added
long val = a[i] % K;
val = (val == 0 ? 0 : K - val);
// Check if the value equals
// to 0 then simply continue
if (val == 0)
continue;
// Check if the value to be
// added is present in the map
if (eqVal.ContainsKey(val))
{
long numVal = eqVal[val];
// Update the answer
maxX = Math.Max(maxX,
val + (K * numVal));
eqVal[val] = 1 +
eqVal.GetValueOrDefault(val, 0);
}
else
{
eqVal.Add(val, 1);
maxX = Math.Max(maxX, val);
}
}
// Print the required result
// We need to add 1 to maxX
// because we cant ignore the
// first move where initially X=0
// and we need to increase it by 1
// to make some changes in array
Console.Write(maxX == 0 ? 0 : maxX + 1);
}
// Driver code
public static void Main(string[] args)
{
int K = 3;
int []a = { 1, 2, 2, 18 };
int N = a.Length;
compute(a, N, K);
}
}
// This code is contributed by rutvik_56
<script>
// Javascript implementation to find the
// minimum number of moves required to
// update the array such that each of
// its element is divisible by K
// Function to find the minimum
// number of moves required to
// update the array such that each of
// its element is divisible by K
function compute(a,N,K)
{
// Initialize Map data structure
let eqVal = new Map();
let maxX = 0;
// Iterate for all the elements
// of the array
for(let i = 0; i < N; i++)
{
// Calculate the
// value to be added
let val = a[i] % K;
val = (val == 0 ? 0 : K - val);
// Check if the value equals
// to 0 then simply continue
if (val == 0)
continue;
// Check if the value to be
// added is present in the map
if (eqVal.has(val))
{
let numVal = eqVal.get(val);
// Update the answer
maxX = Math.max(maxX,
val + (K * numVal));
eqVal.set(val,eqVal.get(val) + 1);
}
else
{
eqVal.set(val, 1);
maxX = Math.max(maxX, val);
}
}
// Print the required result
// We need to add 1 to maxX
// because we cant ignore the
// first move where initially X=0
// and we need to increase it by 1
// to make some changes in array
document.write(maxX == 0 ? 0 : maxX + 1);
}
// Driver code
let K = 3;
let a=[1, 2, 2, 18];
let N = a.length;
compute(a, N, K);
// This code is contributed by avanitrachhadiya2155
</script>
Output:
5
Time Complexity: O(N*logN), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.
