Commit 296536a

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Edit Distance

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	`1`	`+/* Amit Bansal - @amitbansal7 */`
	`2`	`+#include <bits/stdc++.h>`
	`3`	`+using namespace std;`
	`4`	`+`
	`5`	`+/*`
	`6`	`+Edit Distance. Given two text strings A of length n and B of length m, you want to transform A into B`
	`7`	`+with a minimum number of operations of the following types: delete a character from A, insert a`
	`8`	`+character into A, or change some character in A into a new character. The minimal number of such`
	`9`	`+operations required to transform A into B is called the edit distance between A and B.`
	`10`	`+*/`
	`11`	`+`
	`12`	`+int main()`
	`13`	`+{`
	`14`	`+ string a = "FOOD";`
	`15`	`+ string b = "MONEY";`
	`16`	`+`
	`17`	`+ int al = a.length();`
	`18`	`+ int bl = b.length();`
	`19`	`+`
	`20`	`+ int DP[al + 1][bl + 1];`
	`21`	`+`
	`22`	`+ for (int i = 0; i <= al; i++)`
	`23`	`+ {`
	`24`	`+ for (int j = 0; j <= bl; j++)`
	`25`	`+ {`
	`26`	`+ if (i == 0)`
	`27`	`+ DP[i][j] = j;`
	`28`	`+`
	`29`	`+ else if (j == 0)`
	`30`	`+ DP[i][j] = i;`
	`31`	`+`
	`32`	`+ else if (a[i - 1] == b[j - 1])`
	`33`	`+ DP[i][j] = DP[i - 1][j - 1];`
	`34`	`+`
	`35`	`+ else`
	`36`	`+ DP[i][j] = 1 + min(DP[i - 1][j - 1], min(DP[i - 1][j], DP[i][j - 1]));`
	`37`	`+ }`
	`38`	`+ }`
	`39`	`+`
	`40`	`+ printf("%d\n",DP[al][bl]);`
	`41`	`+`
	`42`	`+`
	`43`	`+ return 0;`
	`44`	`+}`

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	`1`	`+/* Amit Bansal - @amitbansal7 */`
	`2`	`+#include <bits/stdc++.h>`
	`3`	`+using namespace std;`
	`4`	`+`
	`5`	`+/*`
	`6`	`+Edit Distance. Given two text strings A of length n and B of length m, you want to transform A into B`
	`7`	`+with a minimum number of operations of the following types: delete a character from A, insert a`
	`8`	`+character into A, or change some character in A into a new character. The minimal number of such`
	`9`	`+operations required to transform A into B is called the edit distance between A and B.`
	`10`	`+*/`
	`11`	`+`
	`12`	`+int DP[20][20];`
	`13`	`+`
	`14`	`+int solve(string a, string b, int i, int j)`
	`15`	`+{`
	`16`	`+ if (i < 1)`
	`17`	`+ return DP[i][j] = j;`
	`18`	`+`
	`19`	`+ else if (j < 1)`
	`20`	`+ return DP[i][j] = i;`
	`21`	`+`
	`22`	`+ else if (a[i-1] == b[j-1])`
	`23`	`+ return DP[i][j] = solve(a, b, i - 1, j - 1);`
	`24`	`+`
	`25`	`+ return DP[i][j] = 1 + min(solve(a, b, i - 1, j - 1),`
	`26`	`+ min(solve(a, b, i - 1, j), solve(a, b, i, j - 1)));`
	`27`	`+}`
	`28`	`+int main()`
	`29`	`+{`
	`30`	`+ string a = "FOOD";`
	`31`	`+ string b = "MONEY";`
	`32`	`+`
	`33`	`+ int al = a.length();`
	`34`	`+ int bl = b.length();`
	`35`	`+`
	`36`	`+ printf("%d\n", solve(a, b, al, bl));`
	`37`	`+`
	`38`	`+ return 0;`
	`39`	`+}`

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