ابحث عن الحد الأدنى لتكلفة التعديل للمصفوفة
جربه على ممارسة GfG 
#practiceLinkDiv { العرض: لا شيء! مهم؛ }
الإخراج
#practiceLinkDiv { العرض: لا شيء! مهم؛ } نظرًا لمصفوفة من الأعداد الصحيحة الموجبة، استبدل كل عنصر في المصفوفة بحيث يكون الفرق بين العناصر المتجاورة في المصفوفة أقل من أو يساوي هدفًا محددًا. نحن بحاجة إلى تقليل تكلفة التعديل التي تمثل مجموع الاختلافات بين القيم الجديدة والقديمة. نحتاج بشكل أساسي إلى تقليل ?|A[i] - A جديد [i]| أين 0 ؟ أنا ؟ n-1 n هو حجم A[] وA جديد [] هي المصفوفة ذات الفرق المجاور أقل من الهدف أو يساويه. افترض أن جميع عناصر المصفوفة أقل من الثابت M = 100.
أمثلة:
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 ابحث عن الحد الأدنى لتكلفة التعديل للمصفوفة جربه!من أجل تقليل تكلفة التعديل ?|A[i] - A جديد [i]| لجميع الفهرس i في المصفوفة |A[i] - A جديد [i]| يجب أن تكون قريبة من الصفر قدر الإمكان. أيضًا |أ[i] - أ جديد [i+1] ]| ؟ هدف.
يمكن حل هذه المشكلة عن طريق البرمجة الديناميكية .دع dp[i] [j] يحدد الحد الأدنى من تكلفة التعديل عند تغيير A[i] إلى j ثم يتم تحديد علاقة DP بواسطة -
dp[i][j] = min{dp[i - 1][k]} + |j - A[i]|
for all k's such that |k - j| ? target
هنا 0 ؟ أنا ؟ ن و 0 ؟ ي ؟ M حيث n هو عدد العناصر في المصفوفة و M = 100. علينا أن نأخذ في الاعتبار كل k بحيث يكون max(j - target 0) ؟ ك ؟ دقيقة (م ي + الهدف)
أخيرًا، سيكون الحد الأدنى لتكلفة تعديل المصفوفة هو min{dp[n - 1][j]} لجميع 0 ? ي ؟ م.الخوارزمية:
- قم بإنشاء مصفوفة ثنائية الأبعاد باستخدام التهيئة dp[n][M+1] لتسجيل أقل تكلفة تعديل لتغيير A[i] إلى j حيث n هو طول المصفوفة وM هي القيمة القصوى لها.
- احسب أصغر تكلفة تعديل لتغيير A[0] إلى j للعنصر الأول في المصفوفة dp[0][j] باستخدام الصيغة dp[0][j] = abs (j - A[0]).
- استبدل A[i] بـ j في عناصر المصفوفة المتبقية dp[i][j] واستخدم الصيغة dp[i][j] = min(dp[i-1][k] + abs(A[i] - j)) حيث تأخذ k جميع القيم الممكنة بين max(j-target0) وmin(Mj+target) للحصول على الحد الأدنى من تكلفة التعديل.
- نظرًا لأن الحد الأدنى لتكلفة التعديل يعطي أقل رقم من الصف الأخير من جدول dp.
وفيما يلي تنفيذ الفكرة المذكورة أعلاه:
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. ?>
الإخراج
Minimum adjustment cost is 75
تعقيد الوقت: يا (ن * م 2 )
المساحة المساعدة: يا (ن * م)
النهج الفعال : تحسين المساحة
في النهج السابق القيمة الحالية موانئ دبي [أنا] [ي] يعتمد فقط على قيم الصف الحالية والسابقة لـ موانئ دبي . لذا، لتحسين تعقيد المساحة، نستخدم مصفوفة أحادية الأبعاد لتخزين الحسابات.
خطوات التنفيذ:
- إنشاء ناقل 1D موانئ دبي من الحجم م+1 .
- قم بتعيين حالة أساسية عن طريق تهيئة قيم موانئ دبي .
- قم الآن بالتكرار على المشكلات الفرعية بمساعدة الحلقة المتداخلة واحصل على القيمة الحالية من الحسابات السابقة.
- الآن قم بإنشاء متجه 1D مؤقت prev_dp تستخدم لتخزين القيم الحالية من الحسابات السابقة.
- بعد كل تكرار قم بتعيين قيمة prev_dp إلى dp لمزيد من التكرار.
- تهيئة متغير الدقة لتخزين الإجابة النهائية وتحديثها بالتكرار عبر Dp.
- في العودة الأخيرة وطباعة الإجابة النهائية المخزنة في الدقة .
تطبيق:
#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
الإخراج
Minimum adjustment cost is 75
تعقيد الوقت: يا (ن * م 2 )
المساحة المساعدة: يا (م)
