Given an integer K and a matrix of N rows and M columns, the task is to find the minimum number of operations required to make all the elements of the matrix equal. In a single operation, K can be added to or subtracted from any element of the matrix. Print -1 if it is impossible to do so.
Examples:
Input: mat[][] = {{2, 4}, {22, 24}}, K = 2
Output: 20
Explanation: mat[0][0] = 2 + (10 * K) = 22 … 10 operations
mat[0][1] = 4 + (9 * K) = 22 … 9 operations
mat[1][0] = 22 … No operation
mat[1][1] = 24 – K = 22 … 1 operations
10 + 9 + 1 = 20Input: mat[][] = {
Explanation: {3, 63, 42},
{18, 12, 12},
{15, 21, 18},
{33, 84, 24}},
K = 3
Output: 63
Approach:
- Since we are only allowed to add or subtract K from any element, we can easily infer that the mod of all the elements with K should be equal because x % K = (x + K) % K = (x – K) % K.
- If that is not the case, simply print -1.
- Otherwise, sort all the elements of the matrix in non-decreasing order and find the median of the sorted elements.
- The minimum number of steps would occur if we convert all the elements to equal to the median.
- Calculate these steps and print the result.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the minimum // number of operations required int minOperations( int n, int m, int k,
vector<vector< int > >& matrix)
{ // Create another array to
// store the elements of matrix
vector< int > arr(n * m, 0);
int mod = matrix[0][0] % k;
for ( int i = 0; i < n; ++i) {
for ( int j = 0; j < m; ++j) {
arr[i * m + j] = matrix[i][j];
// If not possible
if (matrix[i][j] % k != mod) {
return -1;
}
}
}
// Sort the array to get median
sort(arr.begin(), arr.end());
int median = arr[(n * m) / 2];
// To count the minimum operations
int minOperations = 0;
for ( int i = 0; i < n * m; ++i)
minOperations += abs (arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0) {
int median2 = arr[((n * m) / 2) - 1];
int minOperations2 = 0;
for ( int i = 0; i < n * m; ++i)
minOperations2 += abs (arr[i] - median2) / k;
minOperations = min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
} // Driver code int main()
{ vector<vector< int > > matrix = { { 2, 4, 6 },
{ 8, 10, 12 },
{ 14, 16, 18 },
{ 20, 22, 24 } };
int n = matrix.size();
int m = matrix[0].size();
int k = 2;
cout << minOperations(n, m, k, matrix);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG {
// Function to return the minimum
// number of operations required
static int minOperations( int n, int m, int k,
int matrix[][])
{
// Create another array to
// store the elements of matrix
int [] arr = new int [n * m];
int mod = matrix[ 0 ][ 0 ] % k;
for ( int i = 0 ; i < n; ++i) {
for ( int j = 0 ; j < m; ++j) {
arr[i * m + j] = matrix[i][j];
// If not possible
if (matrix[i][j] % k != mod) {
return - 1 ;
}
}
}
// Sort the array to get median
Arrays.sort(arr);
int median = arr[(n * m) / 2 ];
// To count the minimum operations
int minOperations = 0 ;
for ( int i = 0 ; i < n * m; ++i)
minOperations += Math.abs(arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0 ) {
int median2 = arr[((n * m) / 2 ) - 1 ];
int minOperations2 = 0 ;
for ( int i = 0 ; i < n * m; ++i)
minOperations2
+= Math.abs(arr[i] - median2) / k;
minOperations
= Math.min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
}
// Driver code
public static void main(String[] args)
{
int matrix[][] = { { 2 , 4 , 6 },
{ 8 , 10 , 12 },
{ 14 , 16 , 18 },
{ 20 , 22 , 24 } };
int n = matrix.length;
int m = matrix[ 0 ].length;
int k = 2 ;
System.out.println(minOperations(n, m, k, matrix));
}
} // This code is contributed by ihritik |
# Python3 implementation of the approach # Function to return the minimum # number of operations required def minOperations(n, m, k, matrix):
# Create another array to store the
# elements of matrix
arr = [ 0 ] * (n * m)
mod = matrix[ 0 ][ 0 ] % k
for i in range ( 0 , n):
for j in range ( 0 , m):
arr[i * m + j] = matrix[i][j]
# If not possible
if matrix[i][j] % k ! = mod:
return - 1
# Sort the array to get median
arr.sort()
median = arr[(n * m) / / 2 ]
# To count the minimum operations
minOperations = 0
for i in range ( 0 , n * m):
minOperations + = abs (arr[i] - median) / / k
# If there are even elements, then there
# are two medians. We consider the best
# of two as answer.
if (n * m) % 2 = = 0 :
median2 = arr[((n * m) / / 2 ) - 1 ]
minOperations2 = 0
for i in range ( 0 , n * m):
minOperations2 + = abs (arr[i] - median2) / / k
minOperations = min (minOperations,
minOperations2)
# Return minimum operations required
return minOperations
# Driver code if __name__ = = "__main__" :
matrix = [[ 2 , 4 , 6 ],
[ 8 , 10 , 12 ],
[ 14 , 16 , 18 ],
[ 20 , 22 , 24 ]]
n = len (matrix)
m = len (matrix[ 0 ])
k = 2
print (minOperations(n, m, k, matrix))
# This code is contributed by Rituraj Jain |
// C# implementation of the approach using System;
class GFG {
// Function to return the minimum
// number of operations required
static int minOperations( int n, int m, int k,
int [, ] matrix)
{
// Create another array to
// store the elements of matrix
int [] arr = new int [n * m];
int mod = matrix[0, 0] % k;
for ( int i = 0; i < n; ++i) {
for ( int j = 0; j < m; ++j) {
arr[i * m + j] = matrix[i, j];
// If not possible
if (matrix[i, j] % k != mod) {
return -1;
}
}
}
// Sort the array to get median
Array.Sort(arr);
int median = arr[(n * m) / 2];
// To count the minimum operations
int minOperations = 0;
for ( int i = 0; i < n * m; ++i)
minOperations += Math.Abs(arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0) {
int median2 = arr[((n * m) / 2) - 1];
int minOperations2 = 0;
for ( int i = 0; i < n * m; ++i)
minOperations2
+= Math.Abs(arr[i] - median2) / k;
minOperations
= Math.Min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
}
// Driver code
public static void Main()
{
int [, ] matrix = { { 2, 4, 6 },
{ 8, 10, 12 },
{ 14, 16, 18 },
{ 20, 22, 24 } };
int n = matrix.GetLength(0);
int m = matrix.GetLength(1);
int k = 2;
Console.WriteLine(minOperations(n, m, k, matrix));
}
} // This code is contributed by Ryuga |
<script> // Javascript implementation of the approach
// Function to return the minimum
// number of operations required
function minOperations(n, m, k, matrix)
{
// Create another array to
// store the elements of matrix
let arr = new Array(n * m);
arr.fill(0);
let mod = matrix[0][0] % k;
for (let i = 0; i < n; ++i)
{
for (let j = 0; j < m; ++j)
{
arr[i * m + j] = matrix[i][j];
// If not possible
if (matrix[i][j] % k != mod)
{
return -1;
}
}
}
// Sort the array to get median
arr.sort( function (a, b){ return a - b});
let median = arr[parseInt((n * m) / 2, 10)];
// To count the minimum operations
let minOperations = 0;
for (let i = 0; i < n * m; ++i)
minOperations += parseInt(Math.abs(arr[i] - median) / k, 10);
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0)
{
let median2 = arr[parseInt((n * m) / 2, 10)];
let minOperations2 = 0;
for (let i = 0; i < n * m; ++i)
minOperations2 += parseInt(Math.abs(arr[i] - median2) / k, 10);
minOperations = Math.min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
}
// Driver code
let matrix = [ [ 2, 4, 6 ],
[ 8, 10, 12 ],
[ 14, 16, 18 ],
[ 20, 22, 24 ] ];
let n = 4;
let m = 3;
let k = 2;
document.write(minOperations(n, m, k, matrix));
// This code is contributed by mukesh07.
</script> |
36
Minimum operations of given type to make all elements of a Matrix equal with negative numbers
Implementation:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the minimum // number of operations required int minOperations( int n, int m, int k,
vector<vector< int > >& matrix)
{ // Create another array to
// store the elements of matrix
vector< int > arr;
int mod;
// will not work for negative elements, so ..
// adding this
if (matrix[0][0] < 0) {
mod = k - ( abs (matrix[0][0]) % k);
}
else {
mod = matrix[0][0] % k;
}
for ( int i = 0; i < n; ++i) {
for ( int j = 0; j < m; ++j) {
arr.push_back(matrix[i][j]);
// adding this to handle negative elements too .
int val = matrix[i][j];
if (val < 0) {
int res = k - ( abs (val) % k);
if (res != mod) {
return -1;
}
}
else {
int foo = matrix[i][j];
if (foo % k != mod) {
return -1;
}
}
}
}
// Sort the array to get median
sort(arr.begin(), arr.end());
int median = arr[(n * m) / 2];
// To count the minimum operations
int minOperations = 0;
for ( int i = 0; i < n * m; ++i)
minOperations += abs (arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0) {
// changed here as in case of even elements there
// will be 2 medians
int median2 = arr[((n * m) / 2) - 1];
int minOperations2 = 0;
for ( int i = 0; i < n * m; ++i)
minOperations2 += abs (arr[i] - median2) / k;
minOperations = min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
} // Driver code int main()
{ vector<vector< int > > matrix = { { 2, 4, 6 },
{ 8, 10, 12 },
{ 14, 16, 18 },
{ 20, 22, 24 } };
int n = matrix.size();
int m = matrix[0].size();
int k = 2;
cout << minOperations(n, m, k, matrix);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG {
// Function to return the minimum
// number of operations required
public static int minOperations( int n, int m, int k,
int matrix[][])
{
// Create another array to
// store the elements of matrix
Vector<Integer> arr = new Vector<>();
int mod;
// will not work for negative elements, so ..
// adding this
if (matrix[ 0 ][ 0 ] < 0 ) {
mod = k - (Math.abs(matrix[ 0 ][ 0 ]) % k);
}
else {
mod = matrix[ 0 ][ 0 ] % k;
}
for ( int i = 0 ; i < n; ++i) {
for ( int j = 0 ; j < m; ++j) {
arr.add(matrix[i][j]);
// adding this to handle
// negative elements too .
int val = matrix[i][j];
if (val < 0 ) {
int res = k - (Math.abs(val) % k);
if (res != mod) {
return - 1 ;
}
}
else {
int foo = matrix[i][j];
if (foo % k != mod) {
return - 1 ;
}
}
}
}
// Sort the array to get median
Collections.sort(arr);
int median = arr.get((n * m) / 2 );
// To count the minimum operations
int minOperations = 0 ;
for ( int i = 0 ; i < n * m; ++i)
minOperations
+= Math.abs(arr.get(i) - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0 ) {
// changed here as in case of
// even elements there will be 2 medians
int median2 = arr.get(((n * m) / 2 ) - 1 );
int minOperations2 = 0 ;
for ( int i = 0 ; i < n * m; ++i)
minOperations2
+= Math.abs(arr.get(i) - median2) / k;
minOperations
= Math.min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
}
// Driver code
public static void main(String[] args)
{
int matrix[][] = { { 2 , 4 , 6 },
{ 8 , 10 , 12 },
{ 14 , 16 , 18 },
{ 20 , 22 , 24 } };
int n = matrix.length;
int m = matrix[ 0 ].length;
int k = 2 ;
System.out.println(minOperations(n, m, k, matrix));
}
} // This code is contributed by divyesh072019 |
# Python3 implementation of the # above approach # Function to return the minimum # number of operations required def minOperations(n, m, k,
matrix):
# Create another array to
# store the elements of
# matrix
arr = []
# will not work for negative
# elements, so .. adding this
if (matrix[ 0 ][ 0 ] < 0 ):
mod = k - ( abs (matrix[ 0 ][ 0 ]) % k)
else :
mod = matrix[ 0 ][ 0 ] % k
for i in range (n):
for j in range (m):
arr.append(matrix[i][j])
# adding this to handle
# negative elements too .
val = matrix[i][j]
if (val < 0 ):
res = k - ( abs (val) % k)
if (res ! = mod):
return - 1
else :
foo = matrix[i][j]
if (foo % k ! = mod):
return - 1
# Sort the array to get median
arr.sort()
median = arr[(n * m) / / 2 ]
# To count the minimum
# operations
minOperations = 0
for i in range (n * m):
minOperations + = abs (arr[i] -
median) / / k
# If there are even elements,
# then there are two medians.
# We consider the best of two
# as answer.
if ((n * m) % 2 = = 0 ):
# changed here as in case of
# even elements there will be
# 2 medians
median2 = arr[((n * m) / /
2 ) - 1 ]
minOperations2 = 0
for i in range (n * m):
minOperations2 + = abs (arr[i] -
median2) / k
minOperations = min (minOperations,
minOperations2)
# Return minimum operations required
return minOperations
# Driver code if __name__ = = "__main__" :
matrix = [[ 2 , 4 , 6 ],
[ 8 , 10 , 12 ],
[ 14 , 16 , 18 ],
[ 20 , 22 , 24 ]]
n = len (matrix)
m = len (matrix[ 0 ])
k = 2
print (minOperations(n, m, k, matrix))
# This code is contributed by Chitranayal |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG {
// Function to return the minimum
// number of operations required
static int minOperations( int n, int m, int k,
List<List< int > > matrix)
{
// Create another array to
// store the elements of matrix
List< int > arr = new List< int >();
int mod;
// will not work for negative elements, so ..
// adding this
if (matrix[0][0] < 0) {
mod = k - (Math.Abs(matrix[0][0]) % k);
}
else {
mod = matrix[0][0] % k;
}
for ( int i = 0; i < n; ++i) {
for ( int j = 0; j < m; ++j) {
arr.Add(matrix[i][j]);
// adding this to handle negative elements
// too .
int val = matrix[i][j];
if (val < 0) {
int res = k - (Math.Abs(val) % k);
if (res != mod) {
return -1;
}
}
else {
int foo = matrix[i][j];
if (foo % k != mod) {
return -1;
}
}
}
}
// Sort the array to get median
arr.Sort();
int median = arr[(n * m) / 2];
// To count the minimum operations
int minOperations = 0;
for ( int i = 0; i < n * m; ++i)
minOperations += Math.Abs(arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0) {
// changed here as in case of
// even elements there will be 2 medians
int median2 = arr[((n * m) / 2) - 1];
int minOperations2 = 0;
for ( int i = 0; i < n * m; ++i)
minOperations2
+= Math.Abs(arr[i] - median2) / k;
minOperations
= Math.Min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
}
static void Main()
{
List<List< int > > matrix = new List<List< int > >{
new List< int >{ 2, 4, 6 },
new List< int >{ 8, 10, 12 },
new List< int >{ 14, 16, 18 },
new List< int >{ 20, 22, 24 },
};
int n = matrix.Count;
int m = matrix[0].Count;
int k = 2;
Console.Write(minOperations(n, m, k, matrix));
}
} // This code is contributed by divyeshrabadiya07. |
<script> // Javascript implementation of the approach // Function to return the minimum // number of operations required function minOperations(n,m,k,matrix)
{ // Create another array to
// store the elements of matrix
let arr = [];
let mod;
// will not work for negative elements, so ..
// adding this
if (matrix[0][0] < 0)
{
mod = k - (Math.abs(matrix[0][0]) % k);
}
else
{
mod = matrix[0][0] % k;
}
for (let i = 0; i < n; ++i)
{
for (let j = 0; j < m; ++j)
{
arr.push(matrix[i][j]);
// adding this to handle
// negative elements too .
let val = matrix[i][j];
if (val < 0)
{
let res = k - (Math.abs(val) % k);
if (res != mod)
{
return -1;
}
}
else
{
let foo = matrix[i][j];
if (foo % k != mod)
{
return -1;
}
}
}
}
// Sort the array to get median
arr.sort( function (a,b){ return a-b;});
let median = arr[(n * m) / 2];
// To count the minimum operations
let minOperations = 0;
for (let i = 0; i < n * m; ++i)
minOperations += Math.abs(arr[i] - median) / k;
// If there are even elements, then there
// are two medians. We consider the best
// of two as answer.
if ((n * m) % 2 == 0)
{
// changed here as in case of
// even elements there will be 2 medians
let median2 = arr[((n * m) / 2) - 1];
let minOperations2 = 0;
for (let i = 0; i < n * m; ++i)
minOperations2 += Math.abs(arr[i] - median2) / k;
minOperations = Math.min(minOperations, minOperations2);
}
// Return minimum operations required
return minOperations;
} // Driver code let matrix = [[2, 4, 6], [8, 10, 12],
[14, 16, 18],
[20, 22, 24]];
let n = matrix.length; let m = matrix[0].length; let k = 2; document.write(minOperations(n, m, k, matrix)); // This code is contributed by rag2127 </script> |
36
Complexity Analysis:
- Time Complexity: O(n*m* log(n*m)), where n is the number of rows and m is the number of columns of the given matrix.
- Auxiliary Space: O(n*m)
Using Binary Search:
Approach:
We can also use binary search to solve this problem. We can start with the minimum element of the matrix and the maximum element of the matrix as the lower and upper bounds of the binary search. We can then calculate the middle element of the range and count the number of operations required to make all the elements of the matrix equal to the middle element using the given formula. If the number of operations is less than the minimum number of operations seen so far, we update the minimum number of operations. We then update the lower and upper bounds based on whether the middle element is greater than or less than the value we are searching for.
//C++ code for the above approach #include <iostream> #include <vector> #include <algorithm> #include<climits> using namespace std;
int minimumOperations(vector<vector< int >>& mat, int K) {
int n = mat.size();
int m = mat[0].size();
int minVal = INT_MAX;
int maxVal = INT_MIN;
for ( int i = 0; i < n; i++) {
minVal = min(minVal, *min_element(mat[i].begin(), mat[i].end()));
maxVal = max(maxVal, *max_element(mat[i].begin(), mat[i].end()));
}
int minOps = INT_MAX;
while (minVal <= maxVal) {
int midVal = (minVal + maxVal) / 2;
int ops = 2;
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < m; j++) {
if (mat[i][j] < midVal) {
ops += (midVal - mat[i][j]) / K;
} else {
ops += (mat[i][j] - midVal) / K;
}
}
}
if (ops < minOps) {
minOps = ops;
}
if (midVal == maxVal) {
break ;
}
if (ops == (n * m)) {
break ;
}
if (ops < (n * m) / 2) {
maxVal = midVal;
} else {
minVal = midVal + 1;
}
}
return minOps;
} int main() {
vector<vector< int >> mat = { { 2, 4 }, { 22, 24 } };
int K = 2;
cout << minimumOperations(mat, K) << endl;
return 0;
} |
import java.util.*;
class GFG {
static int minimumOperations( int [][] mat, int K) {
int n = mat.length;
int m = mat[ 0 ].length;
int minVal = Integer.MAX_VALUE;
int maxVal = Integer.MIN_VALUE;
for ( int i = 0 ; i < n; i++) {
minVal = Math.min(minVal, Arrays.stream(mat[i]).min().getAsInt());
maxVal = Math.max(maxVal, Arrays.stream(mat[i]).max().getAsInt());
}
int minOps = Integer.MAX_VALUE;
while (minVal <= maxVal) {
int midVal = (minVal + maxVal) / 2 ;
int ops = 2 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < m; j++) {
if (mat[i][j] < midVal) {
ops += (midVal - mat[i][j]) / K;
} else {
ops += (mat[i][j] - midVal) / K;
}
}
}
if (ops < minOps) {
minOps = ops;
}
if (midVal == maxVal) {
break ;
}
if (ops == (n * m)) {
break ;
}
if (ops < (n * m) / 2 ) {
maxVal = midVal;
} else {
minVal = midVal + 1 ;
}
}
return minOps;
}
public static void main(String[] args) {
int [][] mat = { { 2 , 4 }, { 22 , 24 } };
int K = 2 ;
System.out.println(minimumOperations(mat, K));
}
} |
def minimum_operations(mat, K):
n = len (mat)
m = len (mat[ 0 ])
min_val = min (mat[ 0 ])
max_val = max (mat[ - 1 ])
min_ops = float ( 'inf' )
while min_val < = max_val:
mid_val = (min_val + max_val) / / 2
ops = 2
for i in range (n):
for j in range (m):
if mat[i][j] < mid_val:
ops + = (mid_val - mat[i][j]) / / K
else :
ops + = (mat[i][j] - mid_val) / / K
if ops < min_ops:
min_ops = ops
if mid_val = = max_val:
break
if ops = = (n * m):
break
if ops < (n * m) / / 2 :
max_val = mid_val
else :
min_val = mid_val + 1
return min_ops
# Sample Input mat = [[ 2 , 4 ], [ 22 , 24 ]]
K = 2
# Output: 20 print (minimum_operations(mat, K))
|
using System;
class GFG
{ static int MinimumOperations( int [][] mat, int K)
{
int n = mat.Length;
int m = mat[0].Length;
int minVal = int .MaxValue;
int maxVal = int .MinValue;
for ( int i = 0; i < n; i++)
{
minVal = Math.Min(minVal, Array.Find(mat[i], x => x == Array.Find(mat[i], y => true )));
maxVal = Math.Max(maxVal, Array.Find(mat[i], x => x == Array.Find(mat[i], y => true )));
}
int minOps = int .MaxValue;
while (minVal <= maxVal)
{
int midVal = (minVal + maxVal) / 2;
int ops = 2;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < m; j++)
{
if (mat[i][j] < midVal)
{
ops += (midVal - mat[i][j]) / K;
}
else
{
ops += (mat[i][j] - midVal) / K;
}
}
}
if (ops < minOps)
{
minOps = ops;
}
if (midVal == maxVal)
{
break ;
}
if (ops == (n * m))
{
break ;
}
if (ops < (n * m) / 2)
{
maxVal = midVal;
}
else
{
minVal = midVal + 1;
}
}
return minOps;
}
public static void Main( string [] args)
{
int [][] mat = { new int [] { 2, 4 }, new int [] { 22, 24 } };
int K = 2;
Console.WriteLine(MinimumOperations(mat, K));
}
} |
function minimumOperations(mat, K) {
const n = mat.length;
const m = mat[0].length;
let minVal = Math.min(...mat[0]);
let maxVal = Math.max(...mat[n - 1]);
let minOps = Infinity;
while (minVal <= maxVal) {
const midVal = Math.floor((minVal + maxVal) / 2);
let ops = 2;
for (let i = 0; i < n; i++) {
for (let j = 0; j < m; j++) {
if (mat[i][j] < midVal) {
ops += Math.floor((midVal - mat[i][j]) / K);
} else {
ops += Math.floor((mat[i][j] - midVal) / K);
}
}
}
if (ops < minOps) {
minOps = ops;
}
if (midVal === maxVal || ops === n * m) {
break ;
}
if (ops < (n * m) / 2) {
maxVal = midVal;
} else {
minVal = midVal + 1;
}
}
return minOps;
} const mat = [[2, 4], [22, 24]]; const K = 2; console.log(minimumOperations(mat, K)); |
20
Time Complexity: O(n^2 log(max_val – min_val))
Space Complexity: O(1)