Greedy
Array
Two Pointers
Sorting
Description
You are given two integer arrays nums1
and nums2
both of the same length. The advantage of nums1
with respect to nums2
is the number of indices i
for which nums1[i] > nums2[i]
.
Return any permutation of nums1
that maximizes its advantage with respect to nums2
.
Example 1:
Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]
Example 2:
Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]
Constraints:
1 <= nums1.length <= 105
nums2.length == nums1.length
0 <= nums1[i], nums2[i] <= 109
Solutions
Solution 1
Python3 Java C++ Go TypeScript Rust
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15 class Solution :
def advantageCount ( self , nums1 : List [ int ], nums2 : List [ int ]) -> List [ int ]:
nums1 . sort ()
t = sorted (( v , i ) for i , v in enumerate ( nums2 ))
n = len ( nums2 )
ans = [ 0 ] * n
i , j = 0 , n - 1
for v in nums1 :
if v <= t [ i ][ 0 ]:
ans [ t [ j ][ 1 ]] = v
j -= 1
else :
ans [ t [ i ][ 1 ]] = v
i += 1
return ans
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21 class Solution {
public int [] advantageCount ( int [] nums1 , int [] nums2 ) {
int n = nums1 . length ;
int [][] t = new int [ n ][ 2 ] ;
for ( int i = 0 ; i < n ; ++ i ) {
t [ i ] = new int [] { nums2 [ i ] , i };
}
Arrays . sort ( t , ( a , b ) -> a [ 0 ] - b [ 0 ] );
Arrays . sort ( nums1 );
int [] ans = new int [ n ] ;
int i = 0 , j = n - 1 ;
for ( int v : nums1 ) {
if ( v <= t [ i ][ 0 ] ) {
ans [ t [ j --][ 1 ]] = v ;
} else {
ans [ t [ i ++][ 1 ]] = v ;
}
}
return ans ;
}
}
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19 class Solution {
public :
vector < int > advantageCount ( vector < int >& nums1 , vector < int >& nums2 ) {
int n = nums1 . size ();
vector < pair < int , int >> t ;
for ( int i = 0 ; i < n ; ++ i ) t . push_back ({ nums2 [ i ], i });
sort ( t . begin (), t . end ());
sort ( nums1 . begin (), nums1 . end ());
int i = 0 , j = n - 1 ;
vector < int > ans ( n );
for ( int v : nums1 ) {
if ( v <= t [ i ]. first )
ans [ t [ j -- ]. second ] = v ;
else
ans [ t [ i ++ ]. second ] = v ;
}
return ans ;
}
};
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23 func advantageCount ( nums1 [] int , nums2 [] int ) [] int {
n := len ( nums1 )
t := make ([][] int , n )
for i , v := range nums2 {
t [ i ] = [] int { v , i }
}
sort . Slice ( t , func ( i , j int ) bool {
return t [ i ][ 0 ] < t [ j ][ 0 ]
})
sort . Ints ( nums1 )
ans := make ([] int , n )
i , j := 0 , n - 1
for _ , v := range nums1 {
if v <= t [ i ][ 0 ] {
ans [ t [ j ][ 1 ]] = v
j --
} else {
ans [ t [ i ][ 1 ]] = v
i ++
}
}
return ans
}
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20 function advantageCount ( nums1 : number [], nums2 : number []) : number [] {
const n = nums1 . length ;
const idx = Array . from ({ length : n }, ( _ , i ) => i );
idx . sort (( i , j ) => nums2 [ i ] - nums2 [ j ]);
nums1 . sort (( a , b ) => a - b );
const ans = new Array ( n ). fill ( 0 );
let left = 0 ;
let right = n - 1 ;
for ( let i = 0 ; i < n ; i ++ ) {
if ( nums1 [ i ] > nums2 [ idx [ left ]]) {
ans [ idx [ left ]] = nums1 [ i ];
left ++ ;
} else {
ans [ idx [ right ]] = nums1 [ i ];
right -- ;
}
}
return ans ;
}
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21 impl Solution {
pub fn advantage_count ( mut nums1 : Vec < i32 > , nums2 : Vec < i32 > ) -> Vec < i32 > {
let n = nums1 . len ();
let mut idx = ( 0 .. n ). collect :: < Vec < usize >> ();
idx . sort_by ( |& i , & j | nums2 [ i ]. cmp ( & nums2 [ j ]));
nums1 . sort ();
let mut res = vec! [ 0 ; n ];
let mut left = 0 ;
let mut right = n - 1 ;
for & num in nums1 . iter () {
if num > nums2 [ idx [ left ]] {
res [ idx [ left ]] = num ;
left += 1 ;
} else {
res [ idx [ right ]] = num ;
right -= 1 ;
}
}
res
}
}
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