配列の最小調整コストを求める
GfG Practice で試してみる 
#practiceLinkDiv { 表示: なし !重要; }
出力
#practiceLinkDiv { 表示: なし !重要; } 正の整数の配列が指定された場合、配列内の隣接する要素間の差が指定されたターゲット以下になるように、配列内の各要素を置き換えます。新しい値と古い値の差の合計である調整コストを最小限に抑える必要があります。基本的に ?|A[i] - A を最小化する必要があります 新しい [i]|どこに0?私 ？ n-1 n は A[] と A のサイズです 新しい [] は、ターゲット以下の隣接差分を持つ配列です。配列のすべての要素が定数 M = 100 未満であると仮定します。
例:
Input: arr = [1 3 0 3] target = 1
Output: Minimum adjustment cost is 3
Explanation: One of the possible solutions
is [2 3 2 3]
Input: arr = [2 3 2 3] target = 1
Output: Minimum adjustment cost is 0
Explanation: All adjacent elements in the input
array are already less than equal to given target
Input: arr = [55 77 52 61 39 6
25 60 49 47] target = 10
Output: Minimum adjustment cost is 75
Explanation: One of the possible solutions is
[55 62 52 49 39 29 30 40 49 47]
Recommended Practice 配列の最小調整コストを求める 試してみてください!調整コストを最小限に抑えるには ?|A[i] - A 新しい [i]|配列内のすべてのインデックス i について |A[i] - A 新しい [i]|できる限りゼロに近づける必要があります。また |A[i] - A 新しい [i+1] ]| ?ターゲット。
この問題は次の方法で解決できます 動的プログラミング 。dp[i][j] が A[i] を j に変更する際の最小調整コストを定義するとすると、DP 関係は次のように定義されます。
dp[i][j] = min{dp[i - 1][k]} + |j - A[i]|
for all k's such that |k - j| ? target
ここで0?私 ？ nと0?じ？ M ここで、n は配列内の要素の数であり、M = 100 です。max(j - target 0) ? となるようなすべての k を考慮する必要があります。 k ? min(M j + ターゲット)
最後に、配列の最小調整コストは、すべての 0 ? に対して min{dp[n - 1][j]} になります。じ？ M.アルゴリズム：
- 初期化 dp[n][M+1] を使用して 2D 配列を作成し、A[i] を j に変更する最小の調整コストを記録します。ここで、n は配列の長さ、M はその最大値です。
- 式 dp[0][j] = abs (j - A[0]) を使用して、配列 dp[0][j] の最初の要素について A[0] を j に変更する最小の調整コストを計算します。
- 残りの配列要素 dp[i][j] 内の A[i] を j に置き換え、式 dp[i][j] = min(dp[i-1][k] + abs(A[i] - j)) を使用します。ここで、k は max(j-target0) と min(Mj+target) の間の可能なすべての値をとり、最小の調整コストを取得します。
- 最小調整コストとして、dp テーブルの最後の行から最小の数値を与えます。
以下は上記のアイデアの実装です。
C++ // C++ program to find minimum adjustment cost of an array #include using namespace std ; #define M 100 // Function to find minimum adjustment cost of an array int minAdjustmentCost ( int A [] int n int target ) { // dp[i][j] stores minimal adjustment cost on changing // A[i] to j int dp [ n ][ M + 1 ]; // handle first element of array separately for ( int j = 0 ; j <= M ; j ++ ) dp [ 0 ][ j ] = abs ( j - A [ 0 ]); // do for rest elements of the array for ( int i = 1 ; i < n ; i ++ ) { // replace A[i] to j and calculate minimal adjustment // cost dp[i][j] for ( int j = 0 ; j <= M ; j ++ ) { // initialize minimal adjustment cost to INT_MAX dp [ i ][ j ] = INT_MAX ; // consider all k such that k >= max(j - target 0) and // k <= min(M j + target) and take minimum for ( int k = max ( j - target 0 ); k <= min ( M j + target ); k ++ ) dp [ i ][ j ] = min ( dp [ i ][ j ] dp [ i - 1 ][ k ] + abs ( A [ i ] - j )); } } // return minimum value from last row of dp table int res = INT_MAX ; for ( int j = 0 ; j <= M ; j ++ ) res = min ( res dp [ n - 1 ][ j ]); return res ; } // Driver Program to test above functions int main () { int arr [] = { 55 77 52 61 39 6 25 60 49 47 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int target = 10 ; cout < < 'Minimum adjustment cost is ' < < minAdjustmentCost ( arr n target ) < < endl ; return 0 ; }
Java // Java program to find minimum adjustment cost of an array import java.io.* ; import java.util.* ; class GFG { public static int M = 100 ; // Function to find minimum adjustment cost of an array static int minAdjustmentCost ( int A [] int n int target ) { // dp[i][j] stores minimal adjustment cost on changing // A[i] to j int [][] dp = new int [ n ][ M + 1 ] ; // handle first element of array separately for ( int j = 0 ; j <= M ; j ++ ) dp [ 0 ][ j ] = Math . abs ( j - A [ 0 ] ); // do for rest elements of the array for ( int i = 1 ; i < n ; i ++ ) { // replace A[i] to j and calculate minimal adjustment // cost dp[i][j] for ( int j = 0 ; j <= M ; j ++ ) { // initialize minimal adjustment cost to INT_MAX dp [ i ][ j ] = Integer . MAX_VALUE ; // consider all k such that k >= max(j - target 0) and // k <= min(M j + target) and take minimum int k = Math . max ( j - target 0 ); for ( ; k <= Math . min ( M j + target ); k ++ ) dp [ i ][ j ] = Math . min ( dp [ i ][ j ] dp [ i - 1 ][ k ] + Math . abs ( A [ i ] - j )); } } // return minimum value from last row of dp table int res = Integer . MAX_VALUE ; for ( int j = 0 ; j <= M ; j ++ ) res = Math . min ( res dp [ n - 1 ][ j ] ); return res ; } // Driver program public static void main ( String [] args ) { int arr [] = { 55 77 52 61 39 6 25 60 49 47 }; int n = arr . length ; int target = 10 ; System . out . println ( 'Minimum adjustment cost is ' + minAdjustmentCost ( arr n target )); } } // This code is contributed by Pramod Kumar
Python3 # Python3 program to find minimum # adjustment cost of an array M = 100 # Function to find minimum # adjustment cost of an array def minAdjustmentCost ( A n target ): # dp[i][j] stores minimal adjustment # cost on changing A[i] to j dp = [[ 0 for i in range ( M + 1 )] for i in range ( n )] # handle first element # of array separately for j in range ( M + 1 ): dp [ 0 ][ j ] = abs ( j - A [ 0 ]) # do for rest elements # of the array for i in range ( 1 n ): # replace A[i] to j and # calculate minimal adjustment # cost dp[i][j] for j in range ( M + 1 ): # initialize minimal adjustment # cost to INT_MAX dp [ i ][ j ] = 100000000 # consider all k such that # k >= max(j - target 0) and # k <= min(M j + target) and # take minimum for k in range ( max ( j - target 0 ) min ( M j + target ) + 1 ): dp [ i ][ j ] = min ( dp [ i ][ j ] dp [ i - 1 ][ k ] + abs ( A [ i ] - j )) # return minimum value from # last row of dp table res = 10000000 for j in range ( M + 1 ): res = min ( res dp [ n - 1 ][ j ]) return res # Driver Code arr = [ 55 77 52 61 39 6 25 60 49 47 ] n = len ( arr ) target = 10 print ( 'Minimum adjustment cost is' minAdjustmentCost ( arr n target ) sep = ' ' ) # This code is contributed # by sahilshelangia
C# // C# program to find minimum adjustment // cost of an array using System ; class GFG { public static int M = 100 ; // Function to find minimum adjustment // cost of an array static int minAdjustmentCost ( int [] A int n int target ) { // dp[i][j] stores minimal adjustment // cost on changing A[i] to j int [] dp = new int [ n M + 1 ]; // handle first element of array // separately for ( int j = 0 ; j <= M ; j ++ ) dp [ 0 j ] = Math . Abs ( j - A [ 0 ]); // do for rest elements of the array for ( int i = 1 ; i < n ; i ++ ) { // replace A[i] to j and calculate // minimal adjustment cost dp[i][j] for ( int j = 0 ; j <= M ; j ++ ) { // initialize minimal adjustment // cost to INT_MAX dp [ i j ] = int . MaxValue ; // consider all k such that // k >= max(j - target 0) and // k <= min(M j + target) and // take minimum int k = Math . Max ( j - target 0 ); for ( ; k <= Math . Min ( M j + target ); k ++ ) dp [ i j ] = Math . Min ( dp [ i j ] dp [ i - 1 k ] + Math . Abs ( A [ i ] - j )); } } // return minimum value from last // row of dp table int res = int . MaxValue ; for ( int j = 0 ; j <= M ; j ++ ) res = Math . Min ( res dp [ n - 1 j ]); return res ; } // Driver program public static void Main () { int [] arr = { 55 77 52 61 39 6 25 60 49 47 }; int n = arr . Length ; int target = 10 ; Console . WriteLine ( 'Minimum adjustment' + ' cost is ' + minAdjustmentCost ( arr n target )); } } // This code is contributed by Sam007.
JavaScript < script > // Javascript program to find minimum adjustment cost of an array let M = 100 ; // Function to find minimum adjustment cost of an array function minAdjustmentCost ( A n target ) { // dp[i][j] stores minimal adjustment cost on changing // A[i] to j let dp = new Array ( n ); for ( let i = 0 ; i < n ; i ++ ) { dp [ i ] = new Array ( n ); for ( let j = 0 ; j <= M ; j ++ ) { dp [ i ][ j ] = 0 ; } } // handle first element of array separately for ( let j = 0 ; j <= M ; j ++ ) dp [ 0 ][ j ] = Math . abs ( j - A [ 0 ]); // do for rest elements of the array for ( let i = 1 ; i < n ; i ++ ) { // replace A[i] to j and calculate minimal adjustment // cost dp[i][j] for ( let j = 0 ; j <= M ; j ++ ) { // initialize minimal adjustment cost to INT_MAX dp [ i ][ j ] = Number . MAX_VALUE ; // consider all k such that k >= max(j - target 0) and // k <= min(M j + target) and take minimum let k = Math . max ( j - target 0 ); for ( ; k <= Math . min ( M j + target ); k ++ ) dp [ i ][ j ] = Math . min ( dp [ i ][ j ] dp [ i - 1 ][ k ] + Math . abs ( A [ i ] - j )); } } // return minimum value from last row of dp table let res = Number . MAX_VALUE ; for ( let j = 0 ; j <= M ; j ++ ) res = Math . min ( res dp [ n - 1 ][ j ]); return res ; } let arr = [ 55 77 52 61 39 6 25 60 49 47 ]; let n = arr . length ; let target = 10 ; document . write ( 'Minimum adjustment cost is ' + minAdjustmentCost ( arr n target )); // This code is contributed by decode2207. < /script>
PHP // PHP program to find minimum // adjustment cost of an array $M = 100 ; // Function to find minimum // adjustment cost of an array function minAdjustmentCost ( $A $n $target ) { // dp[i][j] stores minimal // adjustment cost on changing // A[i] to j global $M ; $dp = array ( array ()); // handle first element // of array separately for ( $j = 0 ; $j <= $M ; $j ++ ) $dp [ 0 ][ $j ] = abs ( $j - $A [ 0 ]); // do for rest // elements of the array for ( $i = 1 ; $i < $n ; $i ++ ) { // replace A[i] to j and // calculate minimal adjustment // cost dp[i][j] for ( $j = 0 ; $j <= $M ; $j ++ ) { // initialize minimal adjustment // cost to INT_MAX $dp [ $i ][ $j ] = PHP_INT_MAX ; // consider all k such that // k >= max(j - target 0) and // k <= min(M j + target) and // take minimum for ( $k = max ( $j - $target 0 ); $k <= min ( $M $j + $target ); $k ++ ) $dp [ $i ][ $j ] = min ( $dp [ $i ][ $j ] $dp [ $i - 1 ][ $k ] + abs ( $A [ $i ] - $j )); } } // return minimum value // from last row of dp table $res = PHP_INT_MAX ; for ( $j = 0 ; $j <= $M ; $j ++ ) $res = min ( $res $dp [ $n - 1 ][ $j ]); return $res ; } // Driver Code $arr = array ( 55 77 52 61 39 6 25 60 49 47 ); $n = count ( $arr ); $target = 10 ; echo 'Minimum adjustment cost is ' minAdjustmentCost ( $arr $n $target ); // This code is contributed by anuj_67. ?>
出力
Minimum adjustment cost is 75
時間計算量: O(n*m 2 ）
補助スペース: O(n *m)
効率的なアプローチ: スペースの最適化
前のアプローチでは、現在の値 dp[i][j] の現在および前の行の値にのみ依存します。 DP 。したがって、空間の複雑さを最適化するために、単一の 1D 配列を使用して計算を保存します。
実装手順:
- 1D ベクトルを作成する DP サイズの m+1 。
- の値を初期化して基本ケースを設定します。 DP 。
- 次に、入れ子になったループを使用してサブ問題を反復し、以前の計算から現在の値を取得します。
- ここで一時的な 1d ベクトルを作成します 前_dp 以前の計算からの現在の値を保存するために使用されます。
- すべての反復の後、次の値を割り当てます。 前_dp さらに反復するために dp に変換します。
- 変数を初期化する レス 最終的な答えを保存し、Dp を反復処理することでそれを更新します。
- 最後に、に保存されている最終的な答えを返して出力します。 レス 。
実装：
#include using namespace std ; #define M 100 // Function to find minimum adjustment cost of an array int minAdjustmentCost ( int A [] int n int target ) { int dp [ M + 1 ]; // Array to store the minimum adjustment costs for each value for ( int j = 0 ; j <= M ; j ++ ) dp [ j ] = abs ( j - A [ 0 ]); // Initialize the first row with the absolute differences for ( int i = 1 ; i < n ; i ++ ) // Iterate over the array elements { int prev_dp [ M + 1 ]; memcpy ( prev_dp dp sizeof ( dp )); // Store the previous row's minimum costs for ( int j = 0 ; j <= M ; j ++ ) // Iterate over the possible values { dp [ j ] = INT_MAX ; // Initialize the current value with maximum cost // Find the minimum cost by considering the range of previous values for ( int k = max ( j - target 0 ); k <= min ( M j + target ); k ++ ) dp [ j ] = min ( dp [ j ] prev_dp [ k ] + abs ( A [ i ] - j )); } } int res = INT_MAX ; for ( int j = 0 ; j <= M ; j ++ ) res = min ( res dp [ j ]); // Find the minimum cost in the last row return res ; // Return the minimum adjustment cost } int main () { int arr [] = { 55 77 52 61 39 6 25 60 49 47 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int target = 10 ; cout < < 'Minimum adjustment cost is ' < < minAdjustmentCost ( arr n target ) < < endl ; return 0 ; }
Java import java.util.Arrays ; public class MinimumAdjustmentCost { static final int M = 100 ; // Function to find the minimum adjustment cost of an array static int minAdjustmentCost ( int [] A int n int target ) { int [] dp = new int [ M + 1 ] ; // Initialize the first row with absolute differences for ( int j = 0 ; j <= M ; j ++ ) { dp [ j ] = Math . abs ( j - A [ 0 ] ); } // Iterate over the array elements for ( int i = 1 ; i < n ; i ++ ) { int [] prev_dp = Arrays . copyOf ( dp dp . length ); // Store the previous row's minimum costs // Iterate over the possible values for ( int j = 0 ; j <= M ; j ++ ) { dp [ j ] = Integer . MAX_VALUE ; // Initialize the current value with maximum cost // Find the minimum cost by considering the range of previous values for ( int k = Math . max ( j - target 0 ); k <= Math . min ( M j + target ); k ++ ) { dp [ j ] = Math . min ( dp [ j ] prev_dp [ k ] + Math . abs ( A [ i ] - j )); } } } int res = Integer . MAX_VALUE ; for ( int j = 0 ; j <= M ; j ++ ) { res = Math . min ( res dp [ j ] ); // Find the minimum cost in the last row } return res ; // Return the minimum adjustment cost } public static void main ( String [] args ) { int [] arr = { 55 77 52 61 39 6 25 60 49 47 }; int n = arr . length ; int target = 10 ; System . out . println ( 'Minimum adjustment cost is ' + minAdjustmentCost ( arr n target )); } }
Python3 def min_adjustment_cost ( A n target ): M = 100 dp = [ 0 ] * ( M + 1 ) # Initialize the first row of dp with absolute differences for j in range ( M + 1 ): dp [ j ] = abs ( j - A [ 0 ]) # Iterate over the array elements for i in range ( 1 n ): prev_dp = dp [:] # Store the previous row's minimum costs for j in range ( M + 1 ): dp [ j ] = float ( 'inf' ) # Initialize the current value with maximum cost # Find the minimum cost by considering the range of previous values for k in range ( max ( j - target 0 ) min ( M j + target ) + 1 ): dp [ j ] = min ( dp [ j ] prev_dp [ k ] + abs ( A [ i ] - j )) res = float ( 'inf' ) for j in range ( M + 1 ): res = min ( res dp [ j ]) # Find the minimum cost in the last row return res if __name__ == '__main__' : arr = [ 55 77 52 61 39 6 25 60 49 47 ] n = len ( arr ) target = 10 print ( 'Minimum adjustment cost is' min_adjustment_cost ( arr n target ))
C# using System ; class Program { const int M = 100 ; // Function to find minimum adjustment cost of an array static int MinAdjustmentCost ( int [] A int n int target ) { int [] dp = new int [ M + 1 ]; // Array to store the minimum adjustment costs for each value for ( int j = 0 ; j <= M ; j ++ ) { dp [ j ] = Math . Abs ( j - A [ 0 ]); // Initialize the first row with the absolute differences } for ( int i = 1 ; i < n ; i ++ ) // Iterate over the array elements { int [] prevDp = ( int []) dp . Clone (); // Store the previous row's minimum costs for ( int j = 0 ; j <= M ; j ++ ) // Iterate over the possible values { dp [ j ] = int . MaxValue ; // Initialize the current value with maximum cost // Find the minimum cost by considering the range of previous values for ( int k = Math . Max ( j - target 0 ); k <= Math . Min ( M j + target ); k ++ ) { dp [ j ] = Math . Min ( dp [ j ] prevDp [ k ] + Math . Abs ( A [ i ] - j )); } } } int res = int . MaxValue ; for ( int j = 0 ; j <= M ; j ++ ) { res = Math . Min ( res dp [ j ]); // Find the minimum cost in the last row } return res ; // Return the minimum adjustment cost } static void Main () { int [] arr = { 55 77 52 61 39 6 25 60 49 47 }; int n = arr . Length ; int target = 10 ; Console . WriteLine ( 'Minimum adjustment cost is ' + MinAdjustmentCost ( arr n target )); } }
JavaScript const M = 100 ; // Function to find minimum adjustment cost of an array function minAdjustmentCost ( A n target ) { let dp = new Array ( M + 1 ); // Array to store the minimum adjustment costs for each value for ( let j = 0 ; j <= M ; j ++ ) dp [ j ] = Math . abs ( j - A [ 0 ]); // Initialize the first row with the absolute differences for ( let i = 1 ; i < n ; i ++ ) // Iterate over the array elements { let prev_dp = [... dp ]; // Store the previous row's minimum costs for ( let j = 0 ; j <= M ; j ++ ) // Iterate over the possible values { dp [ j ] = Number . MAX_VALUE ; // Initialize the current value with maximum cost // Find the minimum cost by considering the range of previous values for ( let k = Math . max ( j - target 0 ); k <= Math . min ( M j + target ); k ++ ) dp [ j ] = Math . min ( dp [ j ] prev_dp [ k ] + Math . abs ( A [ i ] - j )); } } let res = Number . MAX_VALUE ; for ( let j = 0 ; j <= M ; j ++ ) res = Math . min ( res dp [ j ]); // Find the minimum cost in the last row return res ; // Return the minimum adjustment cost } let arr = [ 55 77 52 61 39 6 25 60 49 47 ]; let n = arr . length ; let target = 10 ; console . log ( 'Minimum adjustment cost is ' + minAdjustmentCost ( arr n target )); // This code is contributed by Kanchan Agarwal
出力
Minimum adjustment cost is 75
時間計算量: O(n*m 2 ）
補助スペース: ああ（男）
